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Ysc
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‎.gitignore

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.DS_Store
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GenerateCodeTable.class
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GenerateCodeTable$TableRow.class
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gh-md-toc
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Solution.class

‎Java/0-Two Sum.java

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E
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1517474043
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tags:
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方法1:
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方法2:
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```
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/*
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Given an array of integers, return indices of the two numbers such that they add up to a specific target.
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You may assume that each input would have exactly one solution, and you may not use the same element twice.
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Example:
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Given nums = [2, 7, 11, 15], target = 9,
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Because nums[0] + nums[1] = 2 + 7 = 9,
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return [0, 1].
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*/
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/*
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Thoughts:
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很简单的一个顺次判断。把j设定成i+1，只检查一轮。
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*/
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class Solution {
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public int[] twoSum(int[] nums, int target) {
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int x[] = new int[2];
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for (int i = 0; i<nums.length; i++){
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for(int j = i+1; j < nums.length; j++){
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if ( target == nums[i] + nums[j]){
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x[0] = i;
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x[1] = j;
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}
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}
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}
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return x;
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}
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}
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/*
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*/
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```

‎Java/1-Reverse Integer.java

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Easy Leetcode 7
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1517474043
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tags:
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方法1:
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pop头部的第一个数字，push到末尾
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pop the first digit of the input number and push it to the end.
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方法2:
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把数字转换成字符串。
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convert the input number into a string.
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注意有个大坑
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重点注意的是：
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Math.abs(Integer.MIN_VALUE); //-2147483648
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Math.abs(Long.MIN_VALUE); //-9223372036854775808
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```
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/*
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*/
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/*
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方法1
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Thoughts:
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参考了别人的答案。
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*/
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class Solution {
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public int reverse(int x) {
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if(x > Integer.MAX_VALUE || x < Integer.MIN_VALUE) return 0;
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long rev = 0;
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int tmp = Math.abs(x); //不取绝对值也可以
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while(tmp != 0){
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rev = (rev * 10) + tmp % 10;
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tmp /= 10;
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}
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if(rev > Integer.MAX_VALUE || rev < Integer.MIN_VALUE) return 0;
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return (int)(x < 0 ? -rev : rev);
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}
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}
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/*
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方法2
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转换成字符串
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!!!注意有个大坑。math.abs
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*/
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class Solution {
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public int reverse(int x) {
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if(x > Integer.MAX_VALUE || x < Integer.MIN_VALUE) return 0;
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long rev = 0;
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String tmp = Long.toString(Math.abs((long)x));
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String reverse = new StringBuffer(tmp).reverse().toString();
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rev = Long.parseLong(reverse);
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if( rev > (long)Integer.MAX_VALUE || rev < (long)Integer.MIN_VALUE
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) return 0;
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return (int)(x < 0 ? -rev : rev);
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}
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}
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```

‎Java/10-Valid Palindrome.java

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Easy Leetcode 125
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1517474043
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tags: String
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##题目不难，但是容易timeexceed。需要考虑如何降低时间复杂度
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方法1:
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自己做的，会超时
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方法2:
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网上的参考，非常直观的方法。把string倒过来
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方法3:
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最主流的方法，two pointers。从两头往中间比较
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方法4:
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也是一个很巧妙的办法，通过建立一个char的array，来avoid大小写的check
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```
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/*
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*/
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/*
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方法1
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Thoughts:
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###
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*/
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class Solution {
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public boolean isNum(char x) {
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if(48<= x && x<=57 ) return true;
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return false;
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}
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public boolean isAlpha(char x) {
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if(65<= x && x<=90 ) return true;
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if(97<= x && x<=122 ) return true;
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return false;
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}
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public boolean isEqual(char x, char y) {
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if(isNum(x)&&isNum(y)){
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return x==y;
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}
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if(isAlpha(x)&&isAlpha(y)){
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if(x ==y || x-y==32 || y-x==32 ){
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return true;
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}
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}
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return false;
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}
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public boolean isPalindrome(String s) {
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boolean flag = true;
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if(s.length() ==0) return true;
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String ss = "";
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for (int i = 0; i < s.length(); i++) {
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char x = s.charAt(i);
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if(isAlpha(x)||isNum(x)) ss += ""+x;
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}
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if(ss.length() == 0 || ss.length() ==1) return true;
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for (int i = 0; i < ss.length()/2; i++) {
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char x = ss.charAt(i);
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char y = ss.charAt(ss.length()-i-1);
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if(!isEqual(x,y)){
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flag = false;
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break;
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}
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}
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return flag;
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}
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}
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/*
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方法2
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非常直观的方法
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*/
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class Solution {
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public boolean isPalindrome(String s) {
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String actual = s.replaceAll("[^A-Za-z0-9]", "").toLowerCase();
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String rev = new StringBuffer(actual).reverse().toString();
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return actual.equals(rev);
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}
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}
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/*
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方法3
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two pointers
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*/
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public class Solution {
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public boolean isPalindrome(String s) {
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if (s.isEmpty()) {
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return true;
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}
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int head = 0, tail = s.length() - 1;
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char cHead, cTail;
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while(head <= tail) {
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cHead = s.charAt(head);
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cTail = s.charAt(tail);
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if (!Character.isLetterOrDigit(cHead)) {
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head++;
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} else if(!Character.isLetterOrDigit(cTail)) {
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tail--;
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} else {
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if (Character.toLowerCase(cHead) != Character.toLowerCase(cTail)) {
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return false;
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}
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head++;
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tail--;
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}
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}
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return true;
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}
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}
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/*
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方法4
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*/
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public class Solution {
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private static final char[]charMap = new char[256];
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static{
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for(int i=0;i<10;i++){
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charMap[i+'0'] = (char)(1+i); // numeric
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}
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for(int i=0;i<26;i++){
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charMap[i+'a'] = charMap[i+'A'] = (char)(11+i); //alphabetic, ignore cases
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}
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}
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public boolean isPalindrome(String s) {
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char[]pChars = s.toCharArray();
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int start = 0,end=pChars.length-1;
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char cS,cE;
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while(start<end){
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cS = charMap[pChars[start]];
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cE = charMap[pChars[end]];
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if(cS!=0 && cE!=0){
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if(cS!=cE)return false;
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start++;
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end--;
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}else{
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if(cS==0)start++;
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if(cE==0)end--;
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}
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}
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return true;
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}
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}
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```

‎Java/12-Reverse Vowels of a String.java

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Easy Leetcode 345
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1517474043
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tags: String
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##题目不难.
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方法1:
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自己做的，会超时
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方法2:
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网上的参考，巧妙地用indexOf来判断是否含有vowel
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```
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/*
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*/
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/*
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方法1
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基本的two pointers
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###
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*/
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import java.util.HashMap;
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import java.util.Map;
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class Solution {
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public String reverseVowels(String s) {
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HashMap<Character,String> vowel = new HashMap();
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vowel.put('a', "1");
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vowel.put('e', "1");
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vowel.put('i', "1");
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vowel.put('o', "1");
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vowel.put('u', "1");
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vowel.put('A', "1");
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vowel.put('E', "1");
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vowel.put('I', "1");
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vowel.put('O', "1");
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vowel.put('U', "1");
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if (s.length() == 0) return "";
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char tmp;
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char[] ss = s.toCharArray();
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int head = 0, tail = ss.length-1;
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for (int i = 0; head < tail; i++) {
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if(vowel.containsKey(ss[head]) &&vowel.containsKey(ss[tail]) ) {
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tmp = ss[head];
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ss[head] = ss[tail];
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ss[tail] = tmp;
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head++;tail--;
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}
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if(vowel.containsKey(ss[head])&& !vowel.containsKey(ss[tail]) ){
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tail--;
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}
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if(!vowel.containsKey(ss[head])&& vowel.containsKey(ss[tail]) ){
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head++;
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}
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if(!vowel.containsKey(ss[head])&& !vowel.containsKey(ss[tail]) ){
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head++;tail--;
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}
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}
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String res = new String(ss);
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return res;
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}
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}
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/*
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方法2
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用indexOf来判断是否含有vowel
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*/
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public class Solution {
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public String reverseVowels(String s) {
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StringBuilder sb = new StringBuilder();
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int j = s.length() - 1;
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for (int i = 0; i < s.length(); i++)
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{
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if ("AEIOUaeiou".indexOf(s.charAt(i)) != -1)
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{
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while (j >= 0 && "AEIOUaeiou".indexOf(s.charAt(j)) == -1)
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{
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j--;
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}
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sb.append(s.charAt(j));
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j--;
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}
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else
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sb.append(s.charAt(i));
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}
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return sb.toString();
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}
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}
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```

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