题目链接 You are given a string s containing lowercase letters and an integer k. You need to : First, change some characters of s to other lowercase English letters. Then divide s into k non-empty disjoint substrings such that each substring is palindrom…
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. For example, given s = "aab",Return [ ["aa","b"], ["a","a","…
题目链接 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. Example: Input: "aab" Output: 1 Explanation: The palindrome partitioning ["aa&…
[抄题]: 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所有可能的回文串分割方案. 给出 s = "aab",返回 [ ["aa", "b"], ["a", "a", "b"] ] [思维问题]: [一句话思路]: [输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): [画图]: [一刷]: 主函数中要记得新建parti…
Atcoder Yet Another Palindrome Partitioning 思路: 一个字符串满足条件的情况是奇数字母个数小于等于1,也就是异或起来是1<<j(0<=j<=25)或者是0 记mark是异或起来的值 状态转移: dp[mark]=dp[mark]+1; dp[mark]=min(dp[mark^(1<<j)]+1,dp[mark]);(0<=j<=25) 注意dp[0]被转移后可能会变成1,但是由它转移的需要dp[0]=0,所以每次…
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",…
题目链接: Lightoj  1044 - Palindrome Partitioning 题目描述: 给一个字符串,问至少分割多少次?分割出来的子串都是回文串. 解题思路: 先把给定串的所有子串是不是回文串处理出来,然后用dp[i] 表示 从起点到串i的位置的最少分割次数,然后结合处理出来的回文串转移一下即可! 还是好蠢哦!自己竟然感觉是一个区间DP,但是n又那么大,完全不是区间DP的作风啊! #include <cmath> #include <cstdio> #include…
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 all possible palindrome partitioning of s. For example, given s = "aab",Return [ ["aa","b"], ["a","a","…
参考:http://www.cppblog.com/wicbnu/archive/2013/03/18/198565.html 我太喜欢用dfs和回溯法了,但是这些暴力的方法加上剪枝之后复杂度依然是很高,显然不能达到题目的要求. 这个时候应该考虑动态规划,并且要复杂度尽量接近O(n^2)的算法. 下面这个方法更加简洁:自长到短找到回文串后,往后dfs,并记录递归深度表示并更新最小划分数.http://fisherlei.blogspot.com/2013/03/leetcode-palindro…