767. Reorganize String

最新推荐文章于 2025-02-03 15:37:10 发布

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最新推荐文章于 2025-02-03 15:37:10 发布

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文章标签： c/c++ runtime

原文链接：http://www.cnblogs.com/ruruozhenhao/p/9884224.html

本文介绍了一种高效的字符串重组算法，该算法能在给定条件下检查并重组字符串，确保相邻字符不相同。通过排序计数的方法，文章详细阐述了如何实现这一目标，并提供了C++代码示例，展示了算法的具体实现过程及性能优势。

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].

Approach #1 Sort by count:

class Solution {
public:
    string reorganizeString(string S) {
        int len = S.length();
        vector<int> count(26, 0);
        string ans = S;
        for (auto s : S) count[s-'a'] += 100;
        for (int i = 0; i < 26; ++i) count[i] += i;
        sort(count.begin(), count.end());
        int start = 1;
        int mid = (len + 1) / 2;
        for (int i = 0; i < 26; ++i) {
            int times = count[i] / 100;
            char c = 'a' + (count[i] % 100);
            if (times > mid) return "";
            for (int j = 0; j < times; ++j) {
                if (start >= len) start = 0;
                ans[start] = c;
                start += 2;
            }
        }
        //ans += '\0';
        return ans;
    }
};

Runtime: 0 ms, faster than 100.00% of C++ online submissions for Reorganize String.

转载于:https://www.cnblogs.com/ruruozhenhao/p/9884224.html