Leetcode-Day Three

1002. Find Common Characters

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:

1 2	Input: ["bella","label","roller"] Output: ["e","l","l"]

Example 2:

1 2	Input: ["cool","lock","cook"] Output: ["c","o"]

Note:

1. 1 <= A.length <= 100
2. 1 <= A[i].length <= 100
3. A[i][j] is a lowercase letter

Solution:

Approach One : 先计算出A[0]字符每个字符出现的次数，然后到其余字符串去匹配，min_times为该字符在各字符串出现的最小次数， min_times = min(min_times,count(temp,A[k]));最小为0，即存在一个字符串不含有该字符，

最终min_times的取值就是该字符要放入result中的次数，解决字符重复出现的问题。

• Time Complexity :O(n^2)
• Space Complexity: O(n)

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int (char c,string str)大专栏  Leetcode-Day Threean> { int result = 0; for(int i = 0;i < str.size();i++) { if(str[i] == c) { result++; } } return result; } vector<string> commonChars(vector<string>& A) { vector<string> result; int min_times = 0; for(int i = 0; i < A[0].size();i++) { char temp = A[0][i]; bool flag = true; min_times = 0; for(int j = 0; j < i;j++) { if(temp == A[0][j]) //判断该字符在前面是否出现过 { flag = false; } } if(flag) //该字符第一次出现 { for(int h = i; h < A[0].size();h++) { if(A[0][h] == temp) { min_times++; } } //得到A[0][i]在第一个字符串出现的次数 times for(int k = 1; k < A.size();k++) { min_times = min(min_times,count(temp,A[k])); } if(min_times != 0) { string str; stringstream stream; stream << temp; str = stream.str(); for(int k = 0;k < min_times;k++) { result.push_back(str); } } } } return result; }