[LeetCode] Scramble String 字符串 dp】的更多相关文章

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…
Scramble String Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the s…
(Version 0.0) 作为一个小弱,这个题目是我第一次碰到三维的动态规划.在自己做的时候意识到了所谓的scramble实际上有两种可能的类型,一类是在较低层的节点进行的两个子节点的对调,这样的情况如果我们从第一层切分点,或者说从较高层的切分点看的话,s1和s2切分点左边的子串所包含的字符的种类个数应该完全一致,同样右边也是完全一致:另一类是在较高层切分点进行的互换,这样我们如果在同层来考察s1和s2的话,会发现s1的切分点左侧的char和s2的切分点右侧的char种类和每种char的数目一…
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 ="great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may cho…
原题地址:https://oj.leetcode.com/problems/scramble-string/ 题意: Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat…
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…
题目链接 利用动态规划的思想, 对于每种状态(i, j)来说都有(i-1, j) 和 (i,j-1) 需要注意的问题 : 初始化的问题,先把i=0和j=0的状态都初始化后才可以进行dp否则发生数组越界 这里学到了一波vector的初始化方法 : 方法 解释 vector< int > v 默认初始化,vector为空, size为0 vector< int > v2(v1) vector v2 = v1 ; v2作为v1的copy vector< int > v = {…
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: k will be a positive integ…
Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…
一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s…
Leetcode之动态规划(DP)专题-712. 两个字符串的最小ASCII删除和(Minimum ASCII Delete Sum for Two Strings) 给定两个字符串s1, s2,找到使两个字符串相等所需删除字符的ASCII值的最小和. 示例 1: 输入: s1 = "sea", s2 = "eat" 输出: 231 解释: 在 "sea" 中删除 "s" 并将 "s" 的值(115)加入总…
Scramble String Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the s…
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" Output: 42 Input: " -42" Output: -42 Input: "4193 with words" Output: 4193 Input: "words and 987" Output: 0 详细分析 这道题的cor…
题意: 判断两个字符串是否互为Scramble字符串,而互为Scramble字符串的定义: 字符串看作是父节点,从字符串某一处切开,生成的两个子串分别是父串的左右子树,再对切开生成的两个子串继续切开,直到无法再切,此时生成为一棵二叉树.对二叉树的任一子树可任意交换其左右分支,如果S1可以通过交换变成S2,则S1,S2互为Scramble字符串. 思路: 对于分割后的子串,应有IsScramble(s1[0,i] , s2[0,i]) && IsSCramble(s1[i,length] ,…
Scramble String Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the s…
Leetcode之动态规划(DP)专题-647. 回文子串(Palindromic Substrings) 给定一个字符串,你的任务是计算这个字符串中有多少个回文子串. 具有不同开始位置或结束位置的子串,即使是由相同的字符组成,也会被计为是不同的子串. 示例 1: 输入: "abc" 输出: 3 解释: 三个回文子串: "a", "b", "c". 示例 2: 输入: "aaa" 输出: 6 说明: 6个回…
Leetcode之动态规划(DP)专题-474. 一和零(Ones and Zeroes) 在计算机界中,我们总是追求用有限的资源获取最大的收益. 现在,假设你分别支配着 m 个 0 和 n 个 1.另外,还有一个仅包含 0 和 1 字符串的数组. 你的任务是使用给定的 m 个 0 和 n 个 1 ,找到能拼出存在于数组中的字符串的最大数量.每个 0 和 1 至多被使用一次. 注意: 给定 0 和 1 的数量都不会超过 100. 给定字符串数组的长度不会超过 600. 示例 1: 输入: Arr…
Leetcode之动态规划(DP)专题-392. 判断子序列(Is Subsequence) 给定字符串 s 和 t ,判断 s 是否为 t 的子序列. 你可以认为 s 和 t 中仅包含英文小写字母.字符串 t 可能会很长(长度 ~= 500,000),而 s 是个短字符串(长度 <=100). 字符串的一个子序列是原始字符串删除一些(也可以不删除)字符而不改变剩余字符相对位置形成的新字符串.(例如,"ace"是"abcde"的一个子序列,而"aec…
Leetcode之动态规划(DP)专题-72. 编辑距离(Edit Distance) 给定两个单词 word1 和 word2,计算出将 word1 转换成 word2 所使用的最少操作数 . 你可以对一个单词进行如下三种操作: 插入一个字符 删除一个字符 替换一个字符 示例 1: 输入: word1 = "horse", word2 = "ros" 输出: 3 解释: horse -> rorse (将 'h' 替换为 'r') rorse -> r…
Scramble String Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the s…
1.3 Given two strings, write a method to decide if one is a permutation of the other. 这道题给定我们两个字符串,让我们判断一个是否为另一个的全排列字符串.在LeetCode中,关于排列的题有如下几道,Permutation Sequence 序列排序,Permutations 全排列, Permutations II 全排列之二 和 Next Permutation 下一个排列.这道题跟它们比起来,算是很简单的…
版权声明: 本文为博主Bravo Yeung(知乎UserName同名)的原创文章,欲转载请先私信获博主允许,转载时请附上网址 http://blog.csdn.net/lzuacm. C#版 - Leetcode 151. 翻转字符串里的单词 - 题解 151.Reverse Words in a String 在线提交: https://leetcode-cn.com/problems/reverse-words-in-a-string/ 题目描述 给定一个字符串,逐个翻转字符串中的每个单词…
版权声明: 本文为博主Bravo Yeung(知乎UserName同名)的原创文章,欲转载请先私信获博主允许,转载时请附上网址 http://blog.csdn.net/lzuacm. Leetcode 557. 反转字符串中的单词 III - 题解 Leetcode 557. Reverse Words in a String III 在线提交: https://leetcode.com/problems/reverse-words-in-a-string-iii/description/ 题…
leetcode87. Scramble String 题意: 给定一个字符串s1,我们可以通过将它分解为两个非空子字符串来表示为二叉树. 思路: 递归解法 对于每对s1,s2. 在s1某处切一刀,s1分成left,right,然后在s2首部开始等长的地方切一刀,切成left,right.只要s1的left和s2的left,s1的right和s2的right同样也构成scramble string就可以了. 递归解法需要剪枝. 剪枝1:s1,s2不等长return 剪枝2:s1,s2字符不相同r…
LeetCode:反转字符串中的元音字母[345] 题目描述 编写一个函数,以字符串作为输入,反转该字符串中的元音字母. 示例 1: 输入: "hello" 输出: "holle" 示例 2: 输入: "leetcode" 输出: "leotcede" 说明:元音字母不包含字母"y". 题目分析 所谓的做题就是把以前背下来的拿过来改一下即可.双指针碰撞模型,之前已经描述过很多次了,此处不在赘述. 知道AEI…
1048. Longest String Chain https://leetcode.com/problems/longest-string-chain/ Let's say word1 is a predecessor of word2 if and only if we can add exactly one letter anywhere in word1 to make it equal to word2.  For example, "abc" is a predecess…
大意: 给定字符串$s$, 长度为$n$, 取$k=\lfloor log2(n)\rfloor$, 第$i$次操作删除一个长度为$2^{i-1}$的子串, 求一种方案使得, $k$次操作后$s$的字典序最小, 输出删除后的字符串. 考虑一些弱化的情况, 每次均删除长为$2$的子串, 共删除$k$次 那么很容易得出$O(n^3)$的$DP$. int n, k; string s, dp[N][N]; int main() { cin>>s>>k; n = s.size(); RE…