[LeetCode] Palindrome Partitioning II

[LeetCode] Palindrome Partitioning II

思路還是采用動態規劃，和WordBreak很像，只不過dict需要事先自己算出，LeetCode的discussion裡有一個方法可以邊動規邊計算dict，本質上沒有區別，兩次循環復雜度為O(n).

public class Solution {
    public int minCut(String s) {
        if (s == null || s.length() < 1) {
            return 0;
        }   
        
        int len = s.length();
        // pal[i][j] == true, means s.substring(i, j + 1) is palindrome
        // obsviously, when j - i < 1, pal[i][j] = true
        boolean pal[][] = getDict(s);
        // dp[i] is the min cuts of s.substring(i, len)
        int dp[] = new int[len];
        dp[len - 1] = 0;
        for (int i = len - 2; i >= 0; i--) {
            dp[i] = len - i - 1; // max cut
            for (int j = i; j < len; j++) {
                if (pal[i][j]) {
                    if (j == len - 1) {
                        dp[i] = 0;
                        break;
                    } else if (dp[j + 1] + 1 < dp[i]){
                        dp[i] = dp[j + 1] + 1;
                    }
                }
            }
        }
        
        return dp[0];
    }
    
    private boolean[][] getDict(String s) {
        int len = s.length();
        boolean[][] pal = new boolean[len][len];
        
        for (int i = len - 1; i >= 0; i--) {
            for (int j = i; j < len; j++) {
                if (s.charAt(i) == s.charAt(j) && (j - i < 2 || pal[i + 1][j - 1])) {
                    pal[i][j] = true;
                }
            }
        }
    
        return pal;
    }
}