Leetcode: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.

分析：这道题可以用动态规划来解。这里介绍两种实现方法，一种是用一个二维bool数组word[i][j]记录s[i,j]是否为palindrome用一个一维int数组f[i]记录s[0,i]的最少cut数。

另一种方法是只用一个二维数组f[i][j]记录s[i,j]最小cut数，因为如果s[i,j]是palindrome那么s[i][j] = 0，所以s[i][j]同时可以用来记录s[i,j]是否为palindrome。

方法一实现：

class Solution {
public:
    int minCut(string s) {
        vector<int> f(s.length()+1,0);
        vector<vector<bool>> words(s.length()+1, vector<bool>(s.length()+1,false));
        for(int i = 0; i < s.length()+1; i++)
            f[i] = i-1;
        for(int i = 0; i < s.length()+1; i++)
            words[i][i] = true;
            
        for(int i = 1; i < s.length() + 1; i++){
            for(int j = i-1; j >= 0; j--){
                if(s[j] == s[i-1] && (i-j <= 2 || words[j+1][i-2])){
                    words[j][i-1] = true;
                    f[i] = min(f[j]+1,f[i]);
                }
            }
        }
        return f[s.length()];
    }
};

方法二实现：

class Solution {
public:
    int minCut(string s) {
        int n = s.length();
        if(n == 0 || n == 1) return 0;
        
        vector<vector<int> > f(n, vector<int>(n, INT_MAX));
        
        for(int i = n-1; i >=0; i--)
            for(int j = i; j < n; j++){
                if(i == j || s[i] == s[j] && (i == j-1 || f[i+1][j-1] == 0)){
                    f[i][j] = 0;
                    if(j < n-1)f[i][n-1] = min(f[i][n-1], f[j+1][n-1]+1);
                }
            }
        
        return f[0][n-1];
    }
};