[LeetCode-132] Palindrome Partitioning II

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.

 

两个DP，第一个获取全部索引i到j是否回文，第二个可以理解为在s[i]到s[j-1]的基础上新增s[j]、s[j]历遍所有组合找到min_cut。

 

class Solution {
public:
    int minCut(string s) {
        int slen = s.length();
        vector<int> min_cut(slen, INT_MAX);
        vector<vector<bool> > is_palin(slen, vector<bool>(slen, false));
        
        // get is palin matrix
        for (int j = 0; j < slen; ++j) {
            for (int i = 0; i <= j; ++i) {
                if (i == j) {
                    is_palin[i][j] = true;
                } else if (i + 1 == j) {
                    is_palin[i][j] = s.at(i) == s.at(j);
                } else {
                    is_palin[i][j] = is_palin[i + 1][j - 1] && s.at(i) == s.at(j);
                }
            }
        }
        
        // get min cut
        int tmp = INT_MAX;
        for (int j = 0; j < slen; ++j) {
            if (is_palin[0][j]) {
                min_cut[j] = 0;
            }
            else {
                tmp = j;
                for (int i = 1; i <= j; ++i) {
                    if (is_palin[i][j]) {
                        tmp = min(tmp, min_cut[i - 1] + 1);
                    }
                }
                min_cut[j] = tmp;
            }
        }
        
        return min_cut[slen - 1];
    }
};