重构

Author: soulmachine

C++/LeetCodet题解(C++版).pdf

@@ -156,12 +156,9 @@ \subsubsection{代码}
         while (!current.empty() && !found) {
             ++level;
             // 先将本层全部置为已访问，防止同层之间互相指向
-            for (auto iter = current.begin(); iter != current.end(); ++iter)
-                visited.insert(*iter);
-            for (auto iter = current.begin(); iter != current.end(); ++iter) {
-            //while (!current.empty()) {
-                const string word = *iter;
-
+            for (auto word : current)
+                visited.insert(word);
+            for (auto word : current) {
                 for (size_t i = 0; i < word.size(); ++i) {
                     string new_word = word;
                     for (char c = 'a'; c <= 'z'; ++c) {
@@ -202,9 +199,8 @@ \subsubsection{代码}
             result.push_back(path);
             reverse(result.back().begin(), result.back().end());
         } else {
-            for (auto iter = father[word].begin(); iter != father[word].end();
-                    ++iter) {
-                buildPath(father, path, start, *iter, result);
+            for (auto f : father[word]) {
+                buildPath(father, path, start, f, result);
             }
         }
         path.pop_back();

C++/chapBFS.tex

@@ -235,8 +235,8 @@ \subsubsection{代码}
         sort(S.begin(), S.end()); // 本题对顺序有要求，需要排序
         vector<int> count(S.back() - S.front() + 1, 0);
         // 计算所有元素的个数
-        for(auto i = S.begin(); i != S.end(); i++) {
-            count[*i - S[0]]++;
+        for (auto i : S) {
+            count[i - S[0]]++;
         }
 
         // 每个元素选择了多少个

C++/chapBruteforce.tex

@@ -165,6 +165,7 @@ \subsubsection{分析}
 
 \subsubsection{代码}
 \begin{Code}
+// LeetCode, Best Time to Buy and Sell Stock
 class Solution {
 public:
     int maxProfit(vector<int> &prices) {
@@ -206,6 +207,7 @@ \subsubsection{分析}
 
 \subsubsection{代码}
 \begin{Code}
+// LeetCode, Best Time to Buy and Sell Stock II
 class Solution {
 public:
     int maxProfit(vector<int> &prices) {

C++/chapGreedy.tex

@@ -27,7 +27,6 @@ \subsubsection{分析}
 \subsubsection{代码}
 \begin{Code}
 //LeetCode, Two Sum
-// 方法1：暴力，O(n^2)
 // 方法2：hash。用一个哈希表，存储每个数对应的下标，复杂度O(n)，代码如下，
 class Solution {
 public:
@@ -571,7 +570,7 @@ \subsubsection{分析}
 \subsubsection{代码}
 
 \begin{Code}
-// LeetCode, Pascal's Triangle
+// LeetCode, Pascal's Triangle II
 // 滚动数组
 class Solution {
 public:

C++/chapImplement.tex

@@ -78,7 +78,7 @@ \subsubsection{代码}
 // 暴力解法，复杂度O(N*M)
 class Solution {
 public:
-    char *strstr(const char *haystack, const char *needle) {
+    char *strStr(const char *haystack, const char *needle) {
         // if needle is empty return the full string
         if (!*needle) return (char*) haystack;
 
@@ -271,11 +271,23 @@ \subsubsection{代码}
 
 \begin{Code}
 // LeetCode, Longest Palindromic Substring
+// 备忘录法，TLE
+typedef string::const_iterator Iterator;
+
+namespace std {
+template<>
+struct hash<pair<Iterator, Iterator>> {
+    size_t operator()(pair<Iterator, Iterator> const& p) const {
+        return ((size_t) &(*p.first)) ^ ((size_t) &(*p.second));
+    }
+};
+}
+
 class Solution {
 public:
     string longestPalindrome(string const& s) {
         cache.clear();
-        return cachedLongestPalindrome(begin(s), end(s));
+        return cachedLongestPalindrome(s.begin(), s.end());
     }
 
 private:

C++/chapString.tex

@@ -153,8 +153,8 @@ \subsubsection{代码}
         }
 
         vector<vector<int> > ret;
-        for(auto iter = tmp_result.begin(); iter != tmp_result.end(); ++iter) {
-            ret.push_back(*iter);
+        for (auto e : tmp_result) {
+            ret.push_back(e);
         }
         return ret;
     }
@@ -524,7 +524,7 @@ \subsubsection{代码}
     bool isSymmetric(TreeNode *left, TreeNode *right) {
         if (!left && !right) return true;   // 终止条件
         if (!left || !right) return false;  // 终止条件
-        return left->val != right->val      // 三方合并
+        return left->val == right->val      // 三方合并
                 && isSymmetric(left->left, right->right)
                 && isSymmetric(left->right, right->left);
     }
@@ -878,11 +878,11 @@ \subsubsection{代码}
         for (int k = start; k <= end; k++) {
             vector<TreeNode*> leftSubs = generate(start, k - 1);
             vector<TreeNode*> rightSubs = generate(k + 1, end);
-            for (auto i = leftSubs.begin(); i < leftSubs.end(); i++) {
-                for (auto j = rightSubs.begin(); j < rightSubs.end(); j++) {
+            for (auto i : leftSubs) {
+                for (auto j : rightSubs) {
                     TreeNode *node = new TreeNode(k);
-                    node->left = *i;
-                    node->right = *j;
+                    node->left = i;
+                    node->right = j;
                     subTree.push_back(node);
                 }
             }
@@ -1483,20 +1483,20 @@ \subsubsection{代码}
 class Solution {
 public:
     int maxPathSum(TreeNode *root) {
-        max = INT_MIN;
+        max_sum = INT_MIN;
         dfs(root);
-        return max;
+        return max_sum;
     }
 private:
-    int max;
+    int max_sum;
     int dfs(const TreeNode *root) {
         if (root == nullptr) return 0;
         int l = dfs(root->left);
         int r = dfs(root->right);
         int sum = root->val;
         if (l > 0) sum += l;
         if (r > 0) sum += r;
-        max = max(max, sum);
+        max_sum = max(max_sum, sum);
         return max(r, l) > 0 ? max(r, l) + root->val : root->val;
     }
 };