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.

动态规划 这一类的题目最重要的就是保护现场。。 不能改变现场的情况：

比如这里 就不能

 if(cur < max){
　　 findmin(s,i + 1,cur + 1);
 }
写成：
if(cur < max){
　　cur ++;
　　pointer = i + 1;
　　findmin(s,pointer,cur);
}
会影响后面的同级别的其他操作。
 1 public class Solution {
 2     int max = 0;
 3     public int minCut(String s) {
 4         // Start typing your Java solution below
 5         // DO NOT write main() function
 6         max = s.length() - 1;
 7         int pointer = 0;
 8         if(s.length() == 0){
 9             return 0;
10         }
11         int cur = 0;
12         findmin(s,pointer,cur);
13         int k = max;
14         return max;
15     }
16     public void findmin(String s,int pointer,int cur){
17         int len = s.length();
18         if(pointer == len){
19             if(cur - 1 < max){
20                 max = cur - 1;
21             }
22         }
23         for(int i = pointer; i < len; i ++){
24             int flag = 0;
25             if(i == pointer){
26                 findmin(s,i + 1,cur + 1);
27             }
28             else if((i - pointer + 1) % 2 == 0){
29                 int mid = (i - pointer - 1)/2;
30                 int j = 0;
31                 for(; j < mid + 1; j ++){
32                     if(s.charAt(j+pointer) != s.charAt(i - j)){
33                         flag = 1;
34                         break;
35                     }
36                 }
37                 if(flag == 0){
38                     if(cur < max){
39                         findmin(s,i + 1,cur + 1);
40                     }
41                 }
42             }
43             else{
44                 int mid = (i - pointer) / 2 - 1;
45                 int j = 0;
46                 for(; j < mid + 1; j ++){
47                     if(s.charAt(j+pointer) != s.charAt(i - j)){
48                         flag = 1;
49                         break;
50                     }
51                 }
52                 if(flag == 0){
53                     if(cur < max){
54                         findmin(s,i + 1,cur + 1);
55                     }
56                 }
57             }
58         }
59     }
60 }

 

 第四遍：

分为两步，第一步找到所有是palindrome的substring。 O(n^2).

第二步是算从0- i,的最小的cut数。1—D dp

dp[i] = 0, if status[0][i] = true(从0到i刚好是个palindrome)

dp[i] = min(dp[k - 1] + 1), if(k 到 i 刚好是palindrome， k 在[1, i]中)

 1 public class Solution {
 2     /*
 3         this is an 1-D DP problem
 4         with an 2-D information 
 5         dp[j] = min(dp[k](i <= k <= j - 1) + 1(if [k, j] is palindrome), 0 (if[i, j] is palindrome));
 6         dp[i] = 0;
 7         i - start point; j - end point.
 8     */
 9     public int minCut(String s) {
10         if(s == null || s.length() == 0) return 0;
11         int len = s.length();
12         int[] dp = new int[len];
13         boolean[][] status = new boolean[len][len];
14         for(int i = 0; i < s.length() * 2; i ++){
15             int left = i / 2;
16             int right = (i + 1) / 2;
17             for(; left > -1 && right < s.length(); left --, right ++){
18                 if(s.charAt(left) == s.charAt(right)){
19                     status[left][right] = true;
20                 }else{
21                     break;
22                 }
23             }
24         }
25         for(int i = 0; i < len; i ++){
26             if(status[0][i] == true) dp[i] = 0;
27             else{
28                 dp[i] = Integer.MAX_VALUE;
29                 for(int j = 1; j <= i; j ++){
30                     if(status[j][i] == true){
31                         dp[i] = Math.min(dp[i], dp[j - 1] + 1);
32                     }
33                 }
34             }
35         }
36         return dp[len - 1];
37     }
38 }

 

posted on 2013-04-16 07:35  Step-BY-Step  阅读(222)  评论(0)  编辑  收藏  举报

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