[LeetCode]132.Palindrome Partitioning II

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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”,”b”] could be produced using 1 cut.

思路

此题可以用动态规划求解。isPal[j][i]表示字符串s的子串s[j…i]是否为回文串，cut[i]表示子串s[0…i]所需要的最小分割数

这里写图片描述

代码

    /**------------------------------------*   日期：2015-03-02*   作者：SJF0115*   题目: 132.Palindrome Partitioning II*   网址：https://oj.leetcode.com/problems/palindrome-partitioning-ii/*   结果：AC*   来源：LeetCode*   博客：---------------------------------------**/#include <iostream>#include <vector>#include <cstring>#include <algorithm>using namespace std;class Solution {public:int minCut(string s) {int size = s.size();if(size == 0){return 0;}//if// isPal[i][j]表示字符串s的子串s[i,j]是否为回文串bool isPal[size][size];memset(isPal,0,sizeof(isPal));// cut[j]表示子串s[0,j]所需要的最小分割数int cut[size];// cut[0,i]for(int i = 0;i < size;++i){// [0,i]最多分割i次cut[i] = i;// 判断s[j,i]是否是回文串for(int j = 0;j <= i;++j){// s[j,i]是回文串if(s[j] == s[i] && (i - j <= 1 || isPal[j+1][i-1])){isPal[j][i] = true;// s[0,i]是回文串if(j == 0){cut[i] = 0;}//ifelse{cut[i] = min(cut[i],cut[j-1]+1);}//else}//if}//for}//forreturn cut[size-1];}};int main(){Solution s;string str("cabababcbc");cout<<s.minCut(str)<<endl;return 0;}