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Description:
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
- A[i][j] is a lowercase letter
题意:给定一个字符串数组S,要求找出数组中所有字符串相同的字符(如果相同的字符有多个也需要找出);
解法:假设最终的结果为res,那么res中的所有字符必然在所有元素中都有出现,也包括第一个字符串;那么,我们可以利用哈希表T存储第一个字符串的字符出现的次数,之后遍历这个字符串数组,为每个字符串都统计其字符出现的次数为哈希表M,此时,只需要修改T中字符出现的次数为Math.min(T中出现的次数,M中出现的次数)即可;最终,T中保留的就是所有字符串相同字符及其出现的次数;
Java
class Solution {
public List<String> commonChars(String[] A) {
Map<Character, Integer> common = new HashMap<>();
frequency(common, A[0]);
for (int i = 1; i < A.length; i++) {
Map<Character, Integer> temp = new HashMap<>();
frequency(temp, A[i]);
for (char ch: common.keySet()) {
common.put(ch, Math.min(common.get(ch), temp.getOrDefault(ch, 0)));
}
}
List<String> res = new ArrayList<>();
for (char ch: common.keySet()) {
while (common.get(ch) > 0) {
res.add("" + ch);
common.put(ch, common.get(ch) - 1);
}
}
return res;
}
private void frequency(Map<Character, Integer> common, String s) {
for (int i = 0; i < s.length(); i++) {
common.put(s.charAt(i), common.getOrDefault(s.charAt(i), 0) + 1);
}
}
}