LeetCode Palindrome Partitioning II

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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.

一般DP，超时了，因为没有将判断回文字的部分优化。

import java.awt.Image;public class Solution {boolean isPalindrome(String s, int u,int v){int head = u;int tail = v;while(head<tail){if(s.charAt(head)!=s.charAt(tail)){return false;}head++;tail--;}return true;}    public int minCut(String s) {                if(s==null)return 0;        if(isPalindrome(s, 0, s.length()-1)){        System.out.println(0);        return 0;    }            int n = s.length();boolean[][] PalindromePair = new boolean [n][n];for(int i=0;i<n;i++){for(int j=i+1;j<n;j++){PalindromePair[i][j]=isPalindrome(s, i, j); }}int [] []DpCount = new int [n][n];for(int i=0;i<n;i++){DpCount[i][i]=0;for(int j=i+1;j<n;j++){if(PalindromePair[i][j]){DpCount[i][j]=0;}else{DpCount[i][j]=j-i;}}}for(int k=1;k<n;k++){for(int i=0;i<n-k;i++){for(int j=i+1;j<k+i;j++){DpCount[i][i+k]=Math.min(DpCount[i][i+k], DpCount[i][j]+DpCount[j+1][i+k]+1);}}}return DpCount[0][n-1];    }        public static void main(String[]args){        Solution solution = new Solution();        String string = new String("aab");        System.out.println(solution.minCut(string));    }}

下面的算法实际上是两个动态规划算法的结合，一个是动态规划求回文字，一个是动态规划求前i个字符有多少个回文字。

DPCount[i]表示子串[0,i]最少切割多少次

public class Solution {    public int minCut(String s) {                if(s==null)return 0;        int n = s.length();        if(n<2)return 0;    boolean[][] PalindromePair = new boolean [n][n];int [] DPCount = new int [n];for(int i=0;i<n;i++){DPCount[i]=i;}for(int i=0;i<n;i++){if(s.charAt(0)==s.charAt(i)&&((i<2)||PalindromePair[1][i-1])){PalindromePair[0][i]=true;DPCount[i]=0;}for(int j=1;j<=i;j++){if(s.charAt(j)==s.charAt(i)&&((i-j<2)||PalindromePair[j+1][i-1])){PalindromePair[j][i]=true;DPCount[i]=Math.min(DPCount[i], DPCount[j-1]+1);}}}return DPCount[n-1];    }        public static void main(String[]args){        Solution solution = new Solution();        String string = new String("aab");        System.out.println(solution.minCut(string));    }}

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