[leetcode] 767. Reorganize String

最新推荐文章于 2019-02-26 16:24:22 发布

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本文介绍了一种用于检查并重组字符串的算法，确保相邻字符不相同。通过统计字符出现次数，并使用特定排序和填充策略实现重组。当某字符出现次数超过一半时，重组无法完成。

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题目：

Given a string S, check if the letters can be rearranged so that two characters that are adjacent to each other are not the same.

If possible, output any possible result. If not possible, return the empty string.

Example 1:

Input: S = "aab"
Output: "aba"

Example 2:

Input: S = "aaab"
Output: ""

Note:

S will consist of lowercase letters and have length in range [1, 500].

代码：

class Solution {
public:
    string reorganizeString(string S) {
        if(S.size() <= 1) return S;
        int chnum[26] = {0};
        for(int i = 0; i < S.size(); i++){
            chnum[S[i]-'a']++;
        }
        for(int i = 0; i < 26; i++){
            chnum[i] = chnum[i] * 100 + i;
        }
        sort(chnum, chnum+26);
        char res[S.size() + 1];
        res[S.size()] = '\0';
        int index = 1;
        for(int i = 0; i < 26; i++){
            char c = chnum[i] % 100 + 'a';
            int count = chnum[i] / 100;
            if(count > (S.size()+1)/2) return "";
            for(int j = 0; j < count; j++){
                if(index >= S.size()){
                    index = 0;
                }
                res[index] = c;
                index += 2;
            }
        }
        return res;
    }
};