1. 题目

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。

返回符合要求的最少分割次数。

示例:
输入: "aab"
输出: 1
解释: 进行一次分割就可将 s 分割成 ["aa","b"] 这样两个回文子串。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/palindrome-partitioning-ii
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

• dp[i]表示到 i 为止的子串最少需要分割多少次
• 如果一个子串为回文串，dp[i] = 0
• 如果不是，遍历所有的 j （j <= i）,如果s[j,i]是回文串，dp[i] = min(dp[i], dp[j-1]+1)

28 / 29 个通过测试用例

# 超时例子
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class Solution {
public:int minCut(string s) {int i,j,n = s.size();vector<int> dp(n,0);if(s.size()<=1)return 0;for(i = 0; i < n; ++i)dp[i] = i;for(i = 1; i < n; ++i){for(j = i; j > 0; --j){if(ispalindrome(s,0,i))dp[i] = 0;else if(ispalindrome(s, j, i))dp[i] = min(dp[i], dp[j-1]+1);}}return dp[n-1];}bool ispalindrome(string& s, int l, int  r){while(l < r){if(s[l++]!=s[r--])return false;}return true;}
};

• 预先预处理得到所有可能的区间是否是是回文串
• 参考：LeetCode 5. 最长回文子串（动态规划）

class Solution {
public:int minCut(string s) {int i,j,len,n = s.size();vector<int> dp(n,0);vector<vector<bool>> ispalind(n,vector<bool>(n,false));if(s.size()<=1)return 0;for(i = 0; i < n; ++i){dp[i] = i;ispalind[i][i] = true;if(i < n-1 && s[i]==s[i+1])ispalind[i][i+1] = true;}for(len = 1; len < n; ++len){for(i = 0; i < n-len; ++i){if(ispalind[i][i+len-1] && i-1>=0 && s[i-1]==s[i+len])//是回文串ispalind[i-1][i+len] = true;}}for(i = 1; i < n; ++i){for(j = i; j > 0; --j){if(ispalind[0][i])dp[i] = 0;else if(ispalind[j][i])dp[i] = min(dp[i], dp[j-1]+1);}}return dp[n-1];}
};

124 ms 7.4 MB

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