1002. Find Common Characters

最新推荐文章于 2022-01-09 17:40:34 发布

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最新推荐文章于 2022-01-09 17:40:34 发布

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文章标签： java runtime

原文链接：https://segmentfault.com/a/1190000018398902

本文介绍了一种算法，用于找出一组字符串中所有共同出现的字符，包括重复字符。通过使用统计表来跟踪每个字符串中字符的频率，然后找到这些字符在所有字符串中出现次数的最小值，从而得出最终的公共字符列表。

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Given an array A of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list (including duplicates). For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.
You may return the answer in any order.
Example 1:

Input: ["bella","label","roller"]
Output: ["e","l","l"]

Example 2:

Input: ["cool","lock","cook"]
Output: ["c","o"]

Note:
1 <= A.length <= 100
1 <= A[i].length <= 100
Ai is a lowercase letter

难度: easy

题目：给定字符串数组Ａ仅由小字符组成，返回所有在所有字符串中都出现过的字符，包括重复。例如，如果一个字符在所有字符串出现了3次而非4次，则返回结果中要包含3次。返回顺序不限。

思路：每个字符串一个统计表。
Runtime: 7 ms, faster than 100.00% of Java online submissions for Find Common Characters.
Memory Usage: 38.1 MB, less than 100.00% of Java online submissions for Find Common Characters.

class Solution {
    public List<String> commonChars(String[] A) {
        int n = A.length;
        int[][] cc = new int[n][26];
        
        for (int i = 0; i < n; i++) {
            for (char c : A[i].toCharArray()) {
                cc[i][c - 'a']++;
            }
        }
        
        List<String> result = new ArrayList<>();
        for (int i = 0; i < 26; i++) {
            int minCount = 100;
            for (int j = 0; j < n; j++) {
                minCount = Math.min(minCount, cc[j][i]);
            }
            
            for (int j = 0; j < minCount; j++) {
                result.add(String.valueOf((char) (i + 'a')));
            }
        }
        
        return result;
    }
}