5/31

Author: sc-yan

Java/String/21-Find Words That Can Be Formed by Characters.java

@@ -0,0 +1,108 @@
+Easy Leetcode 1160
+
+tags: String
+
+方法1:
+自己的方法， 纯intuitive，没有用到复杂的数据结构，就是命令式解法。
+
+方法2:
+自己思考，改进之后的算法。每次都用StringBuilder太消耗资源。把判断contain改成了用hashmap。提高效率
+
+方法3：
+网友的方法。思路是一致的，只不过没有用hashmap，而是用了array去计算个数。map这种计算count类的方法，很多
+都可以转为array。代码没有map清晰，但是效率高了很多。
+方法2用了55ms，方法3才有9s，可见开销差距。
+
+
+```
+
+/*
+方法1
+*/
+
+class Solution {
+    public int countCharacters(String[] words, String chars) {
+        List<String> result = new ArrayList<>();
+        for (String word : words) {
+            boolean flag = true;
+            StringBuilder tmp = new StringBuilder(chars);
+            for (int j = 0; j < word.length(); j++) {
+                Character x = word.charAt(j);
+                int index = tmp.indexOf(x.toString());
+                if ( index>= 0) {
+                    tmp.deleteCharAt(index);
+                } else {
+                    flag = false;
+                }
+            }
+            if (flag) {
+                result.add(word);
+            }
+        }
+        int res = 0;
+        for (String s : result) {
+            res += s.length();
+        }
+        return res;
+    }
+}
+
+/*
+方法2
+*/
+
+public static int countCharacters(String[] words, String chars) {
+        int res = 0;
+        HashMap<Character, Integer> mapA = new HashMap<>();
+        for (int i = 0; i < chars.length(); i++) {
+            char x = chars.charAt(i);
+            mapA.put(x, mapA.getOrDefault(x, 0) + 1);
+        }
+        for (String word : words) {
+            HashMap<Character, Integer> map = (HashMap<Character, Integer>) mapA.clone();
+            boolean flag = true;
+            for (int j = 0; j < word.length(); j++) {
+                char x = word.charAt(j);
+                if ( map.getOrDefault(x, 0) > 0) {
+                    map.put(x, map.get(x) -1);
+                } else {
+                    flag = false;
+                }
+            }
+            if (flag) {
+                res += word.length();
+            }
+        }
+        return res;
+    }
+
+/*
+方法3
+*/
+
+     public static int countCharacters(String[] words, String chars) {
+        int count = 0;
+        int[] seen = new int[26];
+        //Count char of Chars String
+        for (char c : chars.toCharArray())
+            seen[c - 'a']++;
+        // Comparing each word in words
+        for (String word : words) {
+            // simple making copy of seen arr
+            int[] tSeen = Arrays.copyOf(seen, seen.length); 
+            int Tcount = 0;
+            for (char c : word.toCharArray()) {
+                if (tSeen[c - 'a'] > 0) {
+                    tSeen[c - 'a']--;
+                    Tcount++;
+                }
+            }
+            if (Tcount == word.length())
+                count += Tcount;
+        }
+        return count;
+    }
+
+
+
+