leetcode--Palindrome Partitioning II

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.

 

public class Solution {
    public int minCut(String s) {
      　　int cuts = 0;
		int len = s.length();
		if(len > 1){
			boolean[][] isPalindrome = new boolean[len][len];
			//s.substring(i, i + 1) is palindrome (substring with length 1)
			for(int i = 0; i < len; ++i)
				isPalindrome[i][i] = true;
			//substring with length > 1
			for(int i = len - 2; i >= 0; --i){
				for(int j = len - 1; j > i; --j){
					if(j - i == 1)
						isPalindrome[i][j] = (s.charAt(i) == s.charAt(j));
					else
						isPalindrome[i][j] = s.charAt(i) == s.charAt(j) && isPalindrome[i + 1][j - 1];
					isPalindrome[j][i] = isPalindrome[i][j];
				}
			}
			
			int[] minCutting = new int[len];
			for(int i = len - 1; i >= 0; --i){
				if(isPalindrome[i][len - 1])
					minCutting[i] = 0;
				else{
					int min = len + 1;
					for(int j = i + 1 ; j < len; ++j){
						if(isPalindrome[i][j - 1])
							min = (min < 1 + minCutting[j])? min : 1 + minCutting[j];
					}
					minCutting[i] = min;
				}
			}
			cuts = minCutting[0];
		}
		return cuts;
    }
}

　　

转载于:https://www.cnblogs.com/averillzheng/p/3543742.html