LeetCode 132. Palindrome Partitioning II

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class Solution {
public:
    int minCut(string s) {
        const int n = s.size();
        // if s[i~j] is palindrome.
        vector<vector<bool>> valid(n, vector<bool>(n, true));
        // dp[i] = min cuts of s[0~i].
        vector<int> dp(n, n);
        for (int l = 2; l <= n; ++l)
            for (int left = 0, right = left + l - 1; right < n; ++left, ++right)
                valid[left][right] = s[left] == s[right] && valid[left + 1][right - 1];
        for (int right = 0; right < n; ++right) {
            if (valid[0][right]) {
                dp[right] = 0;
                continue;
            }
            for (int left = 0; left < right; ++left)
                if (valid[left + 1][right])
                    dp[right] = min(dp[right], dp[left] + 1);
        }
        return dp[n - 1];
    }
};
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class Solution {
public:
    int minCut(string s) {
        const int n = s.size();
        // dp[i] = min cuts of s[0~i].
        vector<int> dp(n, n);
        for (int m = 0; m < n; ++m) // enumerate middle points.
            for (int d = 0; d <= 1; ++d) // odd and even length palindrome.
                for (int i = m, j = m + d; i >= 0 && j < n && s[i] == s[j]; --i, ++j)
                    dp[j] = min(dp[j], (i ? (dp[i - 1] + 1) : 0));
        return dp[n - 1];
    }
};