Poison

132. Palindrome Partitioning II

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class Solution {
    public int minCut(String s) {
        int n = s.length();

        // 定义 dp[i][j] 为 s[i, j] 是否为回文串，潜在条件：j >= i, dp[i][j] depends on dp[i + 1][j - 1]
        boolean[][] dp = new boolean[n][n];
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                if (s.charAt(i) == s.charAt(j)) {
                    if (i + 1 >= j - 1) {
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1];
                    }
                }
            }
        }

        // 定义 g[i] 为 s[0, i] 字符串分割为回文字符串的最小次数
        int[] g = new int[n];
        g[0] = 0;

        for (int i = 1; i < n; i++) {
            if (dp[0][i]) {
                g[i] = 0;
            } else {
                g[i] = i;
                for (int j = 0; j < i; j++) {
                    if (dp[j + 1][i]) {
                        g[i] = Math.min(g[i], g[j] + 1);
                    }
                }
            }
        }

        return g[n - 1];
    }
}
Reference

132. Palindrome Partitioning II