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You are given an array of strings words (0-indexed).

In one operation, pick two distinct indices i and j, where words[i] is a non-empty string, and move any character from words[i] to any position in words[j].

Return true if you can make every string in words equal using any number of operations, and false otherwise.

Example 1:

Input: words = ["abc","aabc","bc"]Output: trueExplanation: Move the first 'a' in words[1] to the front of words[2],to make words[1] = "abc" and words[2] = "abc".All the strings are now equal to "abc", so return true.

Example 2:

Input: words = ["ab","a"]Output: falseExplanation: It is impossible to make all the strings equal using the operation.

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 100
• words[i] consists of lowercase English letters.

题意：给你一个字符串数组 words（下标 从 0 开始 计数）。在一步操作中，需先选出两个 不同 下标 i 和 j，其中 words[i] 是一个非空字符串，接着将 words[i] 中的 任一 字符移动到 words[j] 中的 任一 位置上。

如果执行任意步操作可以使 words 中的每个字符串都相等，返回 true ；否则，返回 false 。

解法 哈希表

检查每种字符是否可以平均分到所有字符串：

class Solution {
   public:    bool makeEqual(vector
   
    & words) {
           int cnt[26] = {
   0}, n = words.size();        for (int i = 0; i < n; ++i)            for (int j = 0, m = words[i].size(); j < m; ++j) ++cnt[words[i][j] - 'a'];        for (int i = 0; i < 26; ++i)             if (cnt[i] % n != 0) return false;        return true;    }};
   

运行效率如下：

执行用时：8 ms, 在所有 C++ 提交中击败了99.56% 的用户内存消耗：11.2 MB, 在所有 C++ 提交中击败了89.99% 的用户

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