LeetCode 132. Palindrome Partitioning II

原题链接在这里：https://leetcode.com/problems/palindrome-partitioning-ii/

题目：

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.

题解：

求的是至少切几刀能全是palindrome.

Let dp[j] denotes up to i 最少 能分成几块palindrome.

For all 0 <= i <= j 如果[i,j]段是palindrome, 到j点结束那段就可以从 到i点结束那段+1 来跟新最小值.

初始化至少每个字母都分开肯定都是palindrome了.

答案dp[n]-1. 最少分成的块数 比切的刀数 多一.

用二维array来matin [i, j]段是否是palindrome. 两边闭区间，包括i,j对应的char. This could hep reduce running time from O(n^3) to O(n^2).

Time Complexity: O(n^2). n = s.length().

Space: O(n^2).

AC Java:

class Solution {
    public int minCut(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        
        int n = s.length();
        boolean [][] isPal = preCal(s);
        int [] dp = new int[n+1];
        
        for(int j = 0; j<n; j++){
            dp[j+1] = j+1;
            for(int i = 0; i<=j; i++){  //error
                if(isPal[i][j]){
                    dp[j+1] = Math.min(dp[j+1], dp[i]+1);
                }
            }
        }
        
        return dp[n]-1;
    }
    
    private boolean [][] preCal(String s){
        int n = s.length();
        boolean [][] res = new boolean[n][n];
        for(int j = 0; j<n; j++){
            for(int i = j; i>=0; i--){
                if(s.charAt(i) == s.charAt(j) && (j-i<2 || res[i+1][j-1])){     //error
                    res[i][j] = true;
                }
            }
        }
        
        return res;
    }
}

对每一个点按照奇数和偶数两种方式 左右延展若是palindrome就更新dp array. 若不是就停止.

Time Complexity: O(n^2).

Space: O(n).

AC Java: 

class Solution {
    public int minCut(String s) {
        if(s == null || s.length() == 0){
            return 0;
        }
        
        int len = s.length();
        int [] dp = new int[len+1];
        for(int i = 0; i<=len; i++){
            dp[i] = i-1;
        }
        for(int i = 0; i<len; i++){
            // odd length palindrome
            for(int l = i, r = i; l>=0 && r<len && s.charAt(l)==s.charAt(r); l--, r++){
                dp[r+1] = Math.min(dp[r+1], dp[l]+1);
            }
            
            // even length palindrome
            for(int l = i, r = i+1; l>=0 && r<len && s.charAt(l)==s.charAt(r); l--, r++){
                dp[r+1] = Math.min(dp[r+1], dp[l]+1);
            }
        }
        return dp[len];
    }
}

类似Word Break.