131. Palindrome Partitioning substr使用的是坐标值,不使用.begin()..end()这种迭代器 使用dfs,类似于subsets的题,每次判断要不要加入这个数 start每次是起始的位置,判断当前位置到起始位置是不是回文 class Solution { public: vector<vector<string>> partition(string s) { vector<vector<string> > result;…

Palindrome Partitioning II  Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab",Return 1 since the palindrome p…

Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab",Return 1 since the palindrome partitioning ["aa",…

题目: Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab",Return 1 since the palindrome partitioning ["aa&qu…

Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab",Return 1 since the palindrome partitioning ["aa",…

Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab", Return 1 since the palindrome partitioning ["aa"…

求次数的问题一般用DP class Solution(object): def minCut(self, s): """ :type s: str :rtype: int """ n = len(s) maxInt = 2147483647 cuts = [maxInt for x in range(n)] p = self.palinTable(s) for i in range(n): temp = maxInt if p[0][i] ==…

题目链接:https://leetcode-cn.com/problems/palindrome-partitioning-ii/description/ 参考链接:https://blog.csdn.net/jingsuwen1/article/details/51934277 dp[i]存放[0,i)即以前i个字符串的子串的最小切割数,则所求为dp[s.length()]; 前0个字符串和1个字符串不需要切割,所以dp[0]=0,dp[1]=0: 1.初始化:当字串s.substring(0…

https://leetcode.com/problems/palindrome-partitioning-ii/description/ [题意] 给定一个字符串,求最少切割多少下,使得切割后的每个子串都是回文串 [思路] 求一个字符串的所有子串是否是回文串,O(n^2) dp从后往前推 vector<vector<bool> > p(len,vector<bool>(len)); ;i<len-;i++){ ;j<len-;j++){ if(j!=i)…

给定一个字符串 s,将 s 分割成一些子串,使每个子串都是回文串.返回 s 符合要求的的最少分割次数.例如,给出 s = "aab",返回 1 因为进行一次分割可以将字符串 s 分割成 ["aa","b"] 这样两个回文子串.详见:https://leetcode.com/problems/palindrome-partitioning-ii/description/ 首先设置dp变量 cuts[len+1].cuts[i]表示从第i位置到第le…