leetcode — palindrome-partitioning-ii

import java.util.Arrays;

/**
*
* Source : https://oj.leetcode.com/problems/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 PalindromePartition2 {

/**
* 将字符串分割为多个回文字符串的最小分割次数
*
* 定义数组cut[len+1]，cut[i]表示从0-的最小分割数，初始化cut[0] = -1,
* 当i-j是回文字符串的时候，cut[i] = min(cut[i], cut[j]+1)
*
* 最后得出的cut[len+1]就是0-(len+1)之间的最小分割数
*
* @param str
* @return
*/
public int minPartition (String str) {
if (str.length() <= 1) {
return 0;
}
int[] cut = new int[str.length()+1] ;
Arrays.fill(cut, Integer.MAX_VALUE);
boolean[][] table = new boolean[str.length()][str.length()];
cut[0] = -1;
for (int i = str.length()-1; i >= 0; i--) {
for (int j = i; j < str.length(); j++) {
if ((i+1 > j - 1 || table[i+1][j-1]) && str.charAt(i) == str.charAt(j)) {
table[i][j] = true;
}
}
}

for (int i = 1; i <= str.length(); i++) {
for (int j = i-1; j > -1; j--) {
if (table[j][i-1]) {
cut[i] = Math.min(cut[i], cut[j] + 1);
}
}
}

return cut[str.length()];
}

public static void main(String[] args) {
PalindromePartition2 palindromePartition2 = new PalindromePartition2();
System.out.println(palindromePartition2.minPartition("aab") + "==1");
}

}