内容简介: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 includ
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;
}
}
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