Given a non-empty string s and an integer k, rearrange the string such that the same characters are at least distance k from each other.

All input strings are given in lowercase letters. If it is not possible to rearrange the string, return an empty string "".

Example 1:

Input: s = "aabbcc", k = 3
Output: "abcabc"
Explanation: The same letters are at least distance 3 from each other.

Example 2:

Input: s = "aaabc", k = 3
Output: ""
Explanation: It is not possible to rearrange the string.

Example 3:

Input: s = "aaadbbcc", k = 2
Output: "abacabcd"
Explanation: The same letters are at least distance 2 from each other.

Credits:
Special thanks to @elmirap for adding this problem and creating all test cases.

这道题给了我们一个字符串str,和一个整数k,让我们对字符串str重新排序,使得其中相同的字符之间的距离不小于k,这道题的难度标为Hard,看来不是省油的灯。的确,这道题的解法用到了哈希表,堆,和贪婪算法。这道题我最开始想的算法没有通过OJ的大集合超时了,下面的方法是参考网上大神的解法,发现十分的巧妙。我们需要一个哈希表来建立字符和其出现次数之间的映射,然后需要一个堆来保存这每一堆映射,按照出现次数来排序。然后如果堆不为空我们就开始循环,我们找出k和str长度之间的较小值,然后从0遍历到这个较小值,对于每个遍历到的值,如果此时堆为空了,说明此位置没法填入字符了,返回空字符串,否则我们从堆顶取出一对映射,然后把字母加入结果res中,此时映射的个数减1,如果减1后的个数仍大于0,则我们将此映射加入临时集合v中,同时str的个数len减1,遍历完一次,我们把临时集合中的映射对由加入堆中,参见代码如下:

class Solution {
public:
string rearrangeString(string str, int k) {
if (k == ) return str;
string res;
int len = (int)str.size();
unordered_map<char, int> m;
priority_queue<pair<int, char>> q;
for (auto a : str) ++m[a];
for (auto it = m.begin(); it != m.end(); ++it) {
q.push({it->second, it->first});
}
while (!q.empty()) {
vector<pair<int, int>> v;
int cnt = min(k, len);
for (int i = ; i < cnt; ++i) {
if (q.empty()) return "";
auto t = q.top(); q.pop();
res.push_back(t.second);
if (--t.first > ) v.push_back(t);
--len;
}
for (auto a : v) q.push(a);
}
return res;
}
};

类似题目:

Task Scheduler

参考资料:

https://leetcode.com/problems/rearrange-string-k-distance-apart/

https://leetcode.com/discuss/108174/c-unordered_map-priority_queue-solution-using-cache

LeetCode All in One 题目讲解汇总(持续更新中...)

[LeetCode] Rearrange String k Distance Apart 按距离为k隔离重排字符串的更多相关文章

  1. [LeetCode] 358. Rearrange String k Distance Apart 按距离k间隔重排字符串

    Given a non-empty string str and an integer k, rearrange the string such that the same characters ar ...

  2. [LeetCode] 243. Shortest Word Distance 最短单词距离

    Given a list of words and two words word1 and word2, return the shortest distance between these two ...

  3. [Swift]LeetCode358. 按距离为k隔离重排字符串 $ Rearrange String k Distance Apart

    Given a non-empty string str and an integer k, rearrange the string such that the same characters ar ...

  4. Leetcode: Rearrange String k Distance Apart

    Given a non-empty string str and an integer k, rearrange the string such that the same characters ar ...

  5. [leetcode]243. Shortest Word Distance最短单词距离

    Given a list of words and two words word1 and word2, return the shortest distance between these two ...

  6. [leetcode]244. Shortest Word Distance II最短单词距离(允许连环call)

    Design a class which receives a list of words in the constructor, and implements a method that takes ...

  7. [LeetCode] 244. Shortest Word Distance II 最短单词距离 II

    This is a follow up of Shortest Word Distance. The only difference is now you are given the list of ...

  8. [LeetCode] 245. Shortest Word Distance III 最短单词距离 III

    This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as ...

  9. LeetCode 358. Rearrange String k Distance Apart

    原题链接在这里:https://leetcode.com/problems/rearrange-string-k-distance-apart/description/ 题目: Given a non ...

随机推荐

  1. 【NLP】揭秘马尔可夫模型神秘面纱系列文章(四)

    维特比算法解决隐马尔可夫模型解码问题(中文句法标注) 作者:白宁超 2016年7月12日14:08:28 摘要:最早接触马尔可夫模型的定义源于吴军先生<数学之美>一书,起初觉得深奥难懂且无 ...

  2. Write thread-safe servlets [reproduced]

    If you write Web applications in Java, the servlet is your best friend. Whether you write Java Serve ...

  3. CSS知识总结(六)

    CSS常用样式 4.段落样式 1)行高 控制段落内每行高度 line-height : normal | length 例子 源代码: /* CSS代码 */ .normal{ line-height ...

  4. [原创]django+ldap实现单点登录(装饰器和缓存)

    前言 参考本系列之前的文章,我们已经搭建了ldap并且可以通过django来操作ldap了,剩下的就是下游系统的接入了,现在的应用场景,我是分了2个层次,第一层次是统一认证,保证各个系统通过ldap来 ...

  5. EF Code First学习系列

    EF Model First在实际工作中基本用不到,前段时间学了一下,大概的了解一下.现在开始学习Code First这种方式.这也是在实际工作中用到最多的方式. 下面先给出一些目录: 1.什么是Co ...

  6. 《高性能javascript》 领悟随笔之-------DOM编程篇(二)

    <高性能javascript> 领悟随笔之-------DOM编程篇二 序:在javaSctipt中,ECMASCRIPT规定了它的语法,BOM实现了页面与浏览器的交互,而DOM则承载着整 ...

  7. 怎样在Redis通过StackExchange.Redis 存储集合类型List

    StackExchange 是由StackOverFlow出品, 是对Redis的.NET封装,被越来越多的.NET开发者使用在项目中. 绝大部分原先使用ServiceStack的开发者逐渐都转了过来 ...

  8. C++ constructor

    From <<C++ primer>> struct Sales_data { // constructors added Sales_data() = default; Sa ...

  9. Android 手机卫士10--应用管理器

    1.添加不同类型条目 class MyAdapter extends BaseAdapter{ //获取数据适配器中条目类型的总数,修改成两种(纯文本,图片+文字) @Override public ...

  10. 树莓派 Linux备忘

    //更新树莓派 sudo apt-mark hold raspberrypi-bootloader sudo apt-get update sudo apt-get upgrade //配置 rasp ...