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149 lines
3.7 KiB
Markdown
149 lines
3.7 KiB
Markdown
# [767. Reorganize String](https://leetcode.com/problems/reorganize-string/)
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## 题目
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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.
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If possible, output any possible result. If not possible, return the empty string.
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**Example 1**:
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```
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Input: S = "aab"
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Output: "aba"
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```
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**Example 2**:
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```
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Input: S = "aaab"
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Output: ""
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```
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**Note**:
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S will consist of lowercase letters and have length in range [1, 500].
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## 题目大意
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给定一个字符串，要求重新排列字符串，让字符串两两字符不相同，如果可以实现，即输出最终的字符串，如果不能让两两不相同，则输出空字符串。
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## 解题思路
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这道题有 2 种做法。第一种做法是先统计每个字符的出现频率次数，按照频率次数从高往低排序。具体做法就是第 451 题了。如果有一个字母的频次次数超过了 (len(string)+1)/2 那么就返回空字符串。否则输出最终满足题意的字符串。按照频次排序以后，用 2 个指针，一个从 0 开始，另外一个从中间位置开始，依次取出一个字符拼接起来。
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第二种做法是用优先队列，结点是一个结构体，结构体有 2 个字段，一个字段记录是哪个字符，另一个字段记录是这个字符的频次。按照频次的多作为优先级高，用大根堆建立优先队列。注意，这样建立成功的优先队列，重复字母只有一个结点，频次记录在结构体的频次字段中。额外还需要一个辅助队列。优先队列每次都出队一个优先级最高的，然后频次减一，最终结果加上这个字符。然后将这个结点入队。入队的意义是检测这个结点的频次有没有减到 0，如果还不为 0 ，再插入优先队列中。
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```c
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string reorganizeString(string S) {
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vector<int> mp(26);
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int n = S.size();
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for (char c: S)
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++mp[c-'a'];
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priority_queue<pair<int, char>> pq;
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for (int i = 0; i < 26; ++i) {
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if (mp[i] > (n+1)/2) return "";
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if (mp[i]) pq.push({mp[i], i+'a'});
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}
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queue<pair<int, char>> myq;
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string ans;
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while (!pq.empty() || myq.size() > 1) {
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if (myq.size() > 1) { // 注意这里要大于 1，如果是等于 1 的话，频次大的元素一直在输出了，答案就不对了。
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auto cur = myq.front();
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myq.pop();
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if (cur.first != 0) pq.push(cur);
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}
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if (!pq.empty()) {
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auto cur = pq.top();
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pq.pop();
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ans += cur.second;
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cur.first--;
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myq.push(cur);
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}
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}
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return ans;
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}
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```
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## 代码
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```go
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package leetcode
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import (
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"sort"
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)
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func reorganizeString(S string) string {
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fs := frequencySort767(S)
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if fs == "" {
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return ""
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}
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bs := []byte(fs)
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ans := ""
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j := (len(bs)-1)/2 + 1
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for i := 0; i <= (len(bs)-1)/2; i++ {
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ans += string(bs[i])
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if j < len(bs) {
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ans += string(bs[j])
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}
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j++
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}
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return ans
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}
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func frequencySort767(s string) string {
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if s == "" {
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return ""
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}
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sMap := map[byte]int{}
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cMap := map[int][]byte{}
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sb := []byte(s)
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for _, b := range sb {
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sMap[b]++
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if sMap[b] > (len(sb)+1)/2 {
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return ""
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}
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}
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for key, value := range sMap {
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cMap[value] = append(cMap[value], key)
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}
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var keys []int
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for k := range cMap {
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keys = append(keys, k)
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}
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sort.Sort(sort.Reverse(sort.IntSlice(keys)))
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res := make([]byte, 0)
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for _, k := range keys {
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for i := 0; i < len(cMap[k]); i++ {
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for j := 0; j < k; j++ {
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res = append(res, cMap[k][i])
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}
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}
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}
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return string(res)
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}
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``` |