Given a non-empty string, encode the string such that its encoded length is the shortest.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times.

Note:

  1. k will be a positive integer and encoded string will not be empty or have extra space.
  2. You may assume that the input string contains only lowercase English letters. The string's length is at most 160.
  3. If an encoding process does not make the string shorter, then do not
    encode it. If there are several solutions, return any of them is fine.

Example 1:

Input: "aaa"
Output: "aaa"
Explanation: There is no way to encode it such that it is shorter than the input string, so we do not encode it.

Example 2:

Input: "aaaaa"
Output: "5[a]"
Explanation: "5[a]" is shorter than "aaaaa" by 1 character.

Example 3:

Input: "aaaaaaaaaa"
Output: "10[a]"
Explanation: "a9[a]" or "9[a]a" are also valid solutions, both of them have the same length = 5, which is the same as "10[a]".

Example 4:

Input: "aabcaabcd"
Output: "2[aabc]d"
Explanation: "aabc" occurs twice, so one answer can be "2[aabc]d".

Example 5:

Input: "abbbabbbcabbbabbbc"
Output: "2[2[abbb]c]"
Explanation: "abbbabbbc" occurs twice, but "abbbabbbc" can also be encoded to "2[abbb]c", so one answer can be "2[2[abbb]c]".
 
这道题让我们压缩字符串,把相同的字符串用中括号括起来,然后在前面加上出现的次数,感觉还是一道相当有难度的题呢。参考了网上大神的帖子才弄懂该怎么做,这道题还是应该用DP来做。我们建立一个二维的DP数组,其中dp[i][j]表示s在[i, j]范围内的字符串的缩写形式(如果缩写形式长度大于子字符串,那么还是保留子字符串),那么如果s字符串的长度是n,最终我们需要的结果就保存在dp[0][n-1]中,然后我们需要遍历s的所有子字符串,对于任意一段子字符串[i, j],我们\\我们以中间任意位置k来拆分成两段,比较dp[i][k]加上dp[k+1][j]的总长度和dp[i][j]的长度,将长度较小的字符串赋给dp[i][j],然后我们要做的就是在s中取出[i, j]范围内的子字符串t进行合并。合并的方法是我们在取出的字符串t后面再加上一个t,然后在这里面寻找子字符串t的第二个起始位置,如果第二个起始位置小于t的长度的话,说明t包含重复字符串,举个例子吧,比如 t = "abab", 那么t+t = "abababab",我们在里面找第二个t出现的位置为2,小于t的长度4,说明t中有重复出现,重复的个数为t.size()/pos = 2个,那么我们就要把重复的地方放入中括号中,注意中括号里不能直接放这个子字符串,而是应该从dp中取出对应位置的字符串,因为重复的部分有可能已经写成缩写形式了,比如题目中的例子5。再看一个例子,如果t = "abc",那么t+t = "abcabc",我们在里面找第二个t出现的位置为3,等于t的长度3,说明t中没有重复出现,那么replace就还是t。然后我们比较我们得到的replace和dp[i][j]中的字符串长度,把长度较小的赋给dp[i][j]即可,时间复杂度为O(n3),空间复杂度为O(n2),参见代码如下:
 
解法一:
class Solution {
public:
string encode(string s) {
int n = s.size();
vector<vector<string>> dp(n, vector<string>(n, ""));
for (int step = ; step <= n; ++step) {
for (int i = ; i + step - < n; ++i) {
int j = i + step - ;
dp[i][j] = s.substr(i, step);
for (int k = i; k < j; ++k) {
string left = dp[i][k], right = dp[k + ][j];
if (left.size() + right.size() < dp[i][j].size()) {
dp[i][j] = left + right;
}
}
string t = s.substr(i, j - i + ), replace = "";
auto pos = (t + t).find(t, );
if (pos >= t.size()) replace = t;
else replace = to_string(t.size() / pos) + '[' + dp[i][i + pos - ] + ']';
if (replace.size() < dp[i][j].size()) dp[i][j] = replace;
}
}
return dp[][n - ];
}
};

根据热心网友iffalse的留言,我们可以优化上面的方法。如果t是重复的,是不是就不需要再看left.size() + right.size() < dp[i][j].size()了。例如t是abcabcabcabcabc, 最终肯定是5[abc],不需要再看3[abc]+abcabc或者abcabc+3[abc]。对于一个本身就重复的字符串,最小的长度肯定是n[REPEATED],不会是某个left+right。所以应该把k的那个循环放在t和replace那部分代码的后面。这样的确提高了一些运算效率的,参见代码如下:

解法二:

class Solution {
public:
string encode(string s) {
int n = s.size();
vector<vector<string>> dp(n, vector<string>(n, ""));
for (int step = ; step <= n; ++step) {
for (int i = ; i + step - < n; ++i) {
int j = i + step - ;
dp[i][j] = s.substr(i, step);
string t = s.substr(i, j - i + ), replace = "";
auto pos = (t + t).find(t, );
if (pos < t.size()) {
replace = to_string(t.size() / pos) + "[" + dp[i][i + pos - ] + "]";
if (replace.size() < dp[i][j].size()) dp[i][j] = replace;
continue;
}
for (int k = i; k < j; ++k) {
string left = dp[i][k], right = dp[k + ][j];
if (left.size() + right.size() < dp[i][j].size()) {
dp[i][j] = left + right;
}
}
}
}
return dp[][n - ];
}
};

类似题目:

Decode String

Number of Atoms

参考资料:

https://leetcode.com/problems/encode-string-with-shortest-length/

https://leetcode.com/problems/encode-string-with-shortest-length/discuss/95599/Accepted-Solution-in-Java

https://leetcode.com/problems/encode-string-with-shortest-length/discuss/95605/Easy-to-understand-C%2B%2B-O(n3)-solution

https://leetcode.com/problems/encode-string-with-shortest-length/discuss/95619/C%2B%2B-O(N3)-time-O(N2)-space-solution-using-memorized-dynamic-programming-with-detail-explanations

LeetCode All in One 题目讲解汇总(持续更新中...)

[LeetCode] Encode String with Shortest Length 最短长度编码字符串的更多相关文章

  1. Leetcode: Encode String with Shortest Length && G面经

    Given a non-empty string, encode the string such that its encoded length is the shortest. The encodi ...

  2. Leetcode 8. String to Integer (atoi) atoi函数实现 (字符串)

    Leetcode 8. String to Integer (atoi) atoi函数实现 (字符串) 题目描述 实现atoi函数,将一个字符串转化为数字 测试样例 Input: "42&q ...

  3. 【LeetCode每天一题】Length of Last Word(字符串中最后一个单词的长度)

    Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the l ...

  4. [LeetCode] Construct String from Binary Tree 根据二叉树创建字符串

    You need to construct a string consists of parenthesis and integers from a binary tree with the preo ...

  5. [LeetCode] Decode String 解码字符串

    Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where ...

  6. LeetCode : Given a string, find the length of the longest serial substring without repeating characters.

    Given a string, find the length of the longest serial substring without repeating characters. Exampl ...

  7. Leetcode 943. Find the Shortest Superstring(DP)

    题目来源:https://leetcode.com/problems/find-the-shortest-superstring/description/ 标记难度:Hard 提交次数:3/4 代码效 ...

  8. hust--------The Minimum Length (最短循环节)(kmp)

    F - The Minimum Length Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%lld & %l ...

  9. [LeetCode] Reverse String II 翻转字符串之二

    Given a string and an integer k, you need to reverse the first k characters for every 2k characters ...

随机推荐

  1. C#索引器

    索引器允许类或者结构的实例按照与数组相同的方式进行索引取值,索引器与属性类似,不同的是索引器的访问是带参的. 索引器和数组比较: (1)索引器的索引值(Index)类型不受限制 (2)索引器允许重载 ...

  2. BFC的形成条件和特性分析

    初学CSS时,我们学到很多有意思的CSS规则,比如外边距塌陷,还有浮动元素的一些特性等,其实这些规则背后都是BFC这个东西在控制,下面我们来看下BFC到底是什么. 什么是BFC BFC(Block f ...

  3. 用SignalR 2.0开发客服系统[系列2:实现聊天室]

    前言 交流群:195866844 上周发表了 用SignalR 2.0开发客服系统[系列1:实现群发通讯] 这篇文章,得到了很多帮助和鼓励,小弟在此真心的感谢大家的支持.. 这周继续系列2,实现聊天室 ...

  4. java基础练习 字符串,控制流,日历,日期等

    1,对基本控制流程的一些练习 package org.base.practice3; import org.junit.Test; /** * Created with IntelliJ IDEA. ...

  5. JDBC 练习

    建立两个表,一个水果表一个用户表. 1.要求输入账号和密码,登陆成功显示欢迎界面,失败提示错误 2.显示选择界面,输入不同的数字,显示不同的内容,,并实现不同的功能,并返回界面 import java ...

  6. 《分布式事务解决之道》沙龙ppt共享

    大型分布式系统往往由很多“微服务”组成,而不同的微服务往往又连接着不同的数据库,在看似常用的功能背后,可能又需要横跨不同的“微服务”和“数据库”才能实现.那么如何才能保证系统事务的一致性呢?这也同时是 ...

  7. Java程序员应该掌握的10项技能

    这篇文章主要介绍了作为Java程序员应该掌握的10项技能,包括java的知识点与相关的技能,对于java的学习有不错的参考借鉴价值,需要的朋友可以参考下   1.语法:必须比较熟悉,在写代码的时候ID ...

  8. 关于css3的背景渐变

    关于css3的渐变,目前各大浏览器还未做到很好的支持,所以需要在我们使用时加上各大浏览器前缀. -moz-:使用Mozilla内核的浏览器(Firefox浏览器) -webkit-:使用Webkit内 ...

  9. getting started with transformjs

    Introduction In the past two years, more and more friends for mobile web development have used the t ...

  10. jQuery实现DOM加载方法源码分析

    传统的判断dom加载的方法 使用 dom0级 onload事件来进行触发所有浏览器都支持在最初是很流行的写法 我们都熟悉这种写法: window.onload=function(){ ... }  但 ...