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.
Example：
Input: "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.
解释下题目：
给定一个字符串，找出最少划几刀就可以让划分完的子集都是回文数。
1. 动态规划的方法
实际耗时：17ms
public int minCut(String s) {
char[] c = s.toCharArray();
int n = c.length;
int[] cut = new int[n];
boolean[][] pal = new boolean[n][n];
for(int i = 0; i < n; i++) {
int min = i;
for(int j = 0; j <= i; j++) {
if(c[j] == c[i] && (j + 1 > i - 1 || pal[j + 1][i - 1])) {
pal[j][i] = true;
min = j == 0 ? 0 : Math.min(min, cut[j - 1] + 1);
}
}
cut[i] = min;
}
return cut[n - 1];
}
思路是这样的，如果c[j+1]到c[i-1]之间是回文数，而且又有c[j]==c[i]，那么相当于把回文数扩展了。pal[j][i]=true意味着c[j]到c[i]是回文数。
