LeetCode: Palindrome Partitioning II 解题报告

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 }

View Code

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/dp/MinCut_1206.java