[Leetcode 2021 刷题计划] 132. 分割回文串 II

示例 1：
输入：s = "aab"
输出：1
解释：只需一次分割就可将 s 分割成 ["aa","b"] 这样两个回文子串。

示例 2：
输入：s = "a"
输出：0

示例 3：
输入：s = "ab"
输出：1

class Solution {
private:
    vector<vector<int>> dp;
    int minRes = INT_MAX;
    int cur = 0;

    int isPalindrome(string &s, int left, int right) {
        if (left >= right) dp[left][right] = 1;
        if (dp[left][right] == 0) {
            dp[left][right] = s[left] != s[right] ? -1 : isPalindrome(s, left + 1, right - 1);
        }
        return dp[left][right];
    }

    void dfs(string &s, int start) {
        if (start >= s.size()) {
            minRes = min(cur, minRes);
        }
        for (int end = start; end < s.size(); ++end) {
            if(isPalindrome(s, start, end) == 1) {
                ++cur;
                dfs(s, end + 1);
                --cur;
            }
        }
    }
public:
    int minCut(string s) {
        dp.assign(s.size(), vector<int>(s.size()));
        dfs(s, 0);
        return minRes - 1;
    }
};

class Solution {
public:
    int minCut(string s) {
        int n = s.size();
        vector<vector<int>> g(n, vector<int>(n, true));

        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = (s[i] == s[j]) && g[i + 1][j - 1];
            }
        }

        vector<int> f(n, INT_MAX);
        for (int i = 0; i < n; ++i) {
            if (g[0][i]) {
                f[i] = 0;
            } else {
                for (int j = 0; j < i; ++j) {
                    if (g[j + 1][i]) {
                        f[i] = min(f[i], f[j] + 1);
                    }
                }
            }
        }

        return f[n - 1];
    }
};