[LeetCode] Palindrome Partitioning II

最新推荐文章于 2019-06-01 11:25:07 发布

最新推荐文章于 2019-06-01 11:25:07 发布 · 73 阅读

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本文介绍了一种算法，用于计算将字符串分割成多个回文子串所需的最少切割次数。通过动态规划的方法，首先判断字符串中所有子串是否为回文，然后计算从字符串开始到每个位置所需的最少切割数。

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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.

感觉比上面一题难一点。

多一个dp。

res[i] 表示0～i最少需要cut次数

res[i] =min(

1. res[i],

2. if dp[j+1][i] == 1, 1+res[j]

3. else , i-j+res[j]

);

class Solution {
public:
    int minCut(string s) {
        int len = s.size();
        vector<vector<int> > dp(len);
        for (int i = 0; i < len; i++) dp[i].resize(len, -1);
        for (int i = 0; i < len; i++) {
            dp[i][i] = 1;
            if (i < len - 1 && s[i] == s[i+1]) dp[i][i+1] = 1;
        }
        for (int l = 2; l < len; l++) {
            for (int i = 0; i < len; i++) {
                int j = i+l;
                if (j >= len) continue;
                if (dp[i+1][j-1] == 1 && s[i] == s[j]) dp[i][j] = 1;
            }
        }
        
        vector<int> res(len,100000);
        int tmp;
        for (int i = 0; i < len; i++) {
            if (dp[0][i] == 1) res[i] = 0;
            else {
                for (int k = 0; k < i; k++) {
                    if (dp[k+1][i] == 1) tmp = 1 + res[k];
                    else tmp = res[k] + i - k;
                    res[i] = min(res[i], tmp);
                }
            }
        }
        return res[len-1];
    }
};