forked from JojoYang666/Leetcode-1-300
-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy path_132_PalindromePartitioningII.java
55 lines (46 loc) · 1.81 KB
/
_132_PalindromePartitioningII.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
package leetcode_1To300;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _132_PalindromePartitioningII {
/**
* 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.
[][] isPalindrome
[] cuts
i j
abcba s.charAt(i) = s.charAt(j) && isPalindrome[i+1][j-1]
time : O(n^2)
space : O(n^2)
* @param s
* @return
*/
public int minCut(String s) {
if (s == null || s.length() == 0) return 0;
int len = s.length();
int[] cuts = new int[len];
boolean[][] isPalindrome = new boolean[len][len];
for (int i = 0; i < len; i++) {
int min = i;
for (int j = 0; j <= i; j++) {
if (s.charAt(i) == s.charAt(j) && (i - j < 2 || isPalindrome[j + 1][i - 1])) {
isPalindrome[j][i] = true;
min = j == 0 ? 0 : Math.min(min, cuts[j - 1] + 1);
}
}
cuts[i] = min;
}
return cuts[len - 1];
}
}