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.
SOLUTION 1:
使用DP来解决：
1. D[i] 表示前i 个字符切为回文需要的切割数
2. P[i][j]: S.sub(i-j) is a palindrome.
3. 递推公式： D[i] = Math.min(D[i], D[j] + 1), 0 <= j <= i - 1) ，并且要判断 P[j][i - 1]是不是回文。
4. 注意D[0] = -1的用意，它是指当整个字符串判断出是回文是，因为会D[0] + 1 其实应该是结果为0（没有任何切割），所以，应把D[0] 设置为-1
有个转移函数之后，一个问题出现了，就是如何判断[i,j]是否是回文？每次都从i到j比较一遍？太浪费了，这里也是一个DP问题。
定义函数P[i][j] = true if [i,j]为回文那么P[i][j] = str[i] == str[j] && P[i+1][j-1];
1 public class Solution { 2 public int minCut(String s) { 3 if (s == null || s.length() == 0) { 4 return 0; 5 } 6 7 int len = s.length(); 8 9 // D[i]: 前i 个字符切为回文需要的切割数10 int[] D = new int[len + 1];11 D[0] = -1;12 13 // P[i][j]: S.sub(i-j) is a palindrome.14 boolean[][] P = new boolean[len][len];15 16 for (int i = 1; i <= len; i++) {17 // The worst case is cut character one by one.18 D[i] = i - 1;19 for (int j = 0; j <= i - 1; j++) {20 P[j][i - 1] = false;21 if (s.charAt(j) == s.charAt(i - 1) && (i - 1 - j <= 2 || P[j + 1][i - 2])) {22 P[j][i - 1] = true;23 D[i] = Math.min(D[i], D[j] + 1);24 }25 }26 }27 28 return D[len];29 }30 }