【问题描述】[简单]

【解答思路】

计数法

1. minfreq存放最终重复字母的个数 freq存放每次遍历字符串的字母个数

2. minfreq初始化最大值，每遍历一个字符串后，比较minfreq[i]、freq[i]的大小，minfreq[i]更新为两者的最小值。

3. 根据minfreq，输出答案

时间复杂度：O(N^2) 空间复杂度：O(N)

class Solution {public List<String> commonChars(String[] A) {int[] minfreq = new int[26];Arrays.fill(minfreq, Integer.MAX_VALUE);for (String word: A) {int[] freq = new int[26];int length = word.length();for (int i = 0; i < length; ++i) {char ch = word.charAt(i);++freq[ch - 'a'];}for (int i = 0; i < 26; ++i) {minfreq[i] = Math.min(minfreq[i], freq[i]);}}List<String> ans = new ArrayList<String>();for (int i = 0; i < 26; ++i) {for (int j = 0; j < minfreq[i]; ++j) {ans.add(String.valueOf((char) (i + 'a')));}}return ans;}
}

HashMap思路

 public List<String> commonChars(String[] A) {Map<Character, Integer> map = new HashMap<>();if (A.length == 0)return null;for (Character c : A[0].toCharArray())map.put(c, map.getOrDefault(c, 0) + 1);for (int i = 1; i < A.length; i++) {String str = A[i];Map<Character, Integer> map2 = new HashMap<>();for (Character c : str.toCharArray()) {if (map.containsKey(c)) {map2.put(c, Math.min(map2.getOrDefault(c, 0) + 1, map.get(c)));}}map = map2;}List<String> ans = new ArrayList<>();for (Character c : map.keySet()) {int num = map.get(c);for (int i = 0; i < num; i++) {ans.add(String.valueOf(c));}}return ans;

【总结】

1. Java字母题目细节

数组赋值相同的值
Arrays.fill(minfreq, Integer.MAX_VALUE);
char int 互转
char转 int ch - ‘a’
int 转char (char) (i + ‘a’)
int 转char转String
String.valueOf((char) (i + ‘a’))；

2.简单题一点都不简单 ，细节决定成败

参考链接：https://leetcode-cn.com/problems/find-common-characters/solution/cha-zhao-chang-yong-zi-fu-by-leetcode-solution/