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  • 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;
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/averillzheng/p/3543742.html
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