[Leetcode] 132. 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) {
if(s == null || s.length() <= 1) return 0;
boolean[][] isPalindrome = getPalindrome(s);
int[] cut = new int[s.length()];
cut[0] = 0;  //end at 0th index (one letter), zero cut
for(int i = 1; i < s.length(); i++){
if(isPalindrome[0][i]){
cut[i] = 0;
continue;
}
cut[i] = i; //end at ith index (i letters), at most i cuts
for(int j = 0; j < i; j++){
if(isPalindrome[j + 1][i]){
cut[i] = Math.min(cut[i],cut[j] + 1);
}
}
}
return cut[s.length() - 1];
}
private boolean[][] getPalindrome(String s){
boolean[][] result = new boolean[s.length()][s.length()];
for(int i = 0; i < s.length(); i++){
result[i][i] = true;
}
for(int i = 0; i < s.length() - 1; i++){
result[i][i + 1] = (s.charAt(i) == s.charAt(i + 1));
}
for(int length = 2; length < s.length(); length++){
for(int i = 0; i < s.length() - length; i++){
result[i][i + length] = result[i + 1][i + length - 1] && s.charAt(i) == s.charAt(i + length);
}
}
return result;
}
}