1002. Find Common Characters

摘要： 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 (inclu...

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

難度: easy

題目：給定字串陣列Ａ僅由小字元組成，返回所有在所有字串中都出現過的字元，包括重複。例如，如果一個字元在所有字串出現了3次而非4次，則返回結果中要包含3次。返回順序不限。

思路：每個字串一個統計表。

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;
}
}