LeetCode 132. 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 partitioning [“aa”,”b”] could be produced using 1 cut.
题目分析：给出一个字符串，将该字符串分割，使得每部分都是回文串，求最小的额切割次数。

假设dp[i][j]表示s[i…j]的最小切割次数，则有

dp[i][j]=⎧⎩⎨0s[i]!=s[j]min{dp[i][j],dp[i][k]+dp[k+1][j]+1}i=ji+1=ji≤k<j

代码如下：

class Solution {
public:
     int minCut(string s) {
        int n = s.length();
        int dp[n][n];
        memset(dp,0,sizeof(dp));

        for(int i = 0; i < n - 1; i++)
        {
            if(s[i] == s[i+1]) dp[i][i+1] = 0;
            else dp[i][i+1] = 1;
        }

        for(int len = 2; len <= n; len++)
        {
            for(int i = 0; i + len < n; i++)
            {
                int j = i + len;
                dp[i][j] = n;
                if(s[i] == s[j] && dp[i+1][j-1] == 0) dp[i][j] = 0;
                for(int k = i; k < j; k++)
                {
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + 1);
                }
            }
        }


    return dp[0][n-1];
    }
};

结果，超时了。后来看了大神的代码，增加了一个数组，只用了O(n^2)的复杂度。
代码如下：

class Solution {
public:
    int minCut(string s) {
        int n = s.length();
        bool isPal[n][n];
        memset(isPal,0,sizeof(isPal));
        int cut[n];


        for(int j = 0; j < n; j++) {
            cut[j] = j;
            for(int i = 0; i <= j; i++) {
                if(s[i] == s[j] && (j - i <= 1 || isPal[i+1][j-1] == true) ) {
                    isPal[i][j] = true;

                    if(i > 0) {
                        cut[j] = min(cut[j],cut[i-1] + 1);
                    }
                    else {
                        cut[j] = 0;
                    }
                }
            }

        }

        return cut[n-1];
    }
};

对于每个cut[j],同样是将[0…j]分成两部分，只不过，其中一部分是回文子串。
这道题和Leetcode 279. Perfect Squares的技巧很相似。