LeetCode 132. Palindrome Partitioning II

132. Palindrome Partitioning II

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• Total Accepted: 65178
• Total Submissions: 276173
• Difficulty: Hard
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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.

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【题目分析】

给定一个字符串，把该字符串切成几段，每段都是一个回文字符串，那么最少的切割次数是多少次？

【思路】

1. 递归

a. 如果当前字符串是回文串，则返回0；

b. 否则找到以第一个字符串开头的回文字符串，该子字符串使得剩下的字符串的切割数最少。

2. 动态规划

先把代码贴下面，但是还没有理解 discuss

【java代码——递归】该算法会导致超时

public class Solution {
    public int minCut(String s) {
        if(isPalindrome(s)) return 0;
        int res = s.length()-1;
        
        for(int i = 1; i < s.length(); i++) {
            if(!isPalindrome(s.substring(0,i))) continue;
            res = Math.min(res, minCut(s.substring(i))+1);
        }
        
        return res;
    }
    
    public boolean isPalindrome(String s) {
        if(s.length() <= 1) return true;
        int left = 0, right = s.length()-1;
        while(left < right) {
            if(s.charAt(left++) != s.charAt(right--)) return false;
        }
        
        return true;
    }
}

【c++代码2——动态规划】

class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        vector<int> cut(n+1, 0);  // number of cuts for the first k characters
        for (int i = 0; i <= n; i++) cut[i] = i-1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; i-j >= 0 && i+j < n && s[i-j]==s[i+j] ; j++) // odd length palindrome
                cut[i+j+1] = min(cut[i+j+1],1+cut[i-j]);

            for (int j = 1; i-j+1 >= 0 && i+j < n && s[i-j+1] == s[i+j]; j++) // even length palindrome
                cut[i+j+1] = min(cut[i+j+1],1+cut[i-j+1]);
        }
        return cut[n];
    }
};