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.

本来是按照Palindrome Partitioning那题的思路改的，复杂度是O(n^3)吧，但是TLE了，考虑降低复杂度，算法中可以降低复杂度的只有判断字符串是否是回文那部分，因为每次计算没有考虑之前计算过的情况，将这部分提出来，用动态规划去做，isPa[i][j] = s.charAt(i) == s.charAt(j) && (i == j || i + 1 == j || isPa[i + 1][j - 1])，这样，判断回文的复杂度为O(n^2)，然后自底向上计算时复杂度为O(n^2)，总体的复杂度为O(n^2)。代码如下：

 1 public int minCut(String s) {
 2         boolean isPa[][] = new boolean[s.length()][s.length()];
 3         if (s == null || s.equals("")) {
 4             return 0;
 5         }
 6         for (int i = s.length() - 1; i >= 0; i--) {
 7             for (int j = i; j < s.length(); j++) {
 8                 isPa[i][j] = s.charAt(i) == s.charAt(j)
 9                         && (i == j || i + 1 == j || isPa[i + 1][j - 1]);
10             }
11         }
12         int result[] = new int[s.length()];
13         for (int i = 0; i < s.length(); i++) {
14             result[i] = 100000;
15         }
16         for (int i = s.length() - 1; i >= 0; i--) {
17             for (int j = i; j < s.length(); j++) {
18                 if (isPa[i][j]) {
19                     if (j == s.length() - 1) {
20                         result[i] = 0;
21                     } else {
22                         result[i] = result[i] < result[j + 1] + 1 ? result[i]
23                                 : result[j + 1] + 1;
24                     }
25                 }
26             }
27         }
28         return result[0];
29     }