1002. Find Common Characters

本题难度并不大，我第一次见也解了出来，也跟最优解非常近似，只是使用了一堆无用的HashMap，实际上本题只需要两个HashMap:

• 记录全局的minimum occurence
  
• 记录单个String的 minimum occurence，然后每个String遍历完后跟全局更新一下即可
  

代码如下：

class Solution {
  public List<String> commonChars(String[] A) {
    int len = A.length;
    int[] minCount = new int[26];
    Arrays.fill(minCount, Integer.MAX_VALUE);
    for (String s : A) {
      int[] count = new int[26];
      s.chars().forEach(ch -> count[ch-'a']++);
      for (int i = 0; i < 26; ++i) {
        minCount[i] = Math.min(minCount[i], count[i]);
      }
    }
    List<String> result = new ArrayList<>();
    for (int i = 0; i < 26; ++i) {
      if (minCount[i] == 0) {
        continue;
      }
      String[] chArr = new String[minCount[i]];
      Arrays.fill(chArr, Character.toString('a' + i));
      result.addAll(Arrays.asList(chArr));
    }
    return result;
  }
}