Palindrome Partitioning II 改进

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.

 维护两张表，一张一维的表是为了从后往前计算递归 以减小开销。

另外一张是 用在计算每一步的大小。（计算的很巧妙）

 1 public class Solution {
 2      boolean[][] store;
 3             int[] results;
 4             int len;
 5             public int minCut(String s) {
 6                 // Start typing your Java solution below
 7                 // DO NOT write main() function
 8                 len = s.length();
 9                 store = new boolean[len][len];
10                 results = new int[len];
11                 getStore(s);
12                 for(int pos = len - 1; pos > -1; pos --){
13                     finmin(s,pos);
14                 }
15                 return results[0];
16             }
17             public void getStore(String s){
18                 int header = 0;
19                 int tail = 0;
20                 for(int i = 0; i < len; i ++){
21                     store[i][i] = true;
22                 }
23                 // for odd length string
24                 for(int i = 1; i < len - 1; i ++){
25                     int k = 1;
26                     while(i + k < len && i - k > -1){
27                         if(s.charAt(i - k) == s.charAt(i + k)){
28                             store[i - k][i + k] = true;
29                         }
30                         else{
31                             break;
32                         }
33                         k ++;
34                     }
35                 }
36                 //for even length string
37                 for(int i = 0; i < len - 1; i ++){
38                     if(s.charAt(i) == s.charAt(i + 1)){
39                         store[i][i + 1] = true;
40                         int k = 1;
41                         while(i + 1 + k < len && i - k > -1){
42                             if(s.charAt(i - k) == s.charAt(i + 1 + k)){
43                                 store[i - k][i + 1 + k] = true;
44                             }
45                             else{
46                                 break;
47                             }
48                             k ++;
49                         }
50                     }
51                 }
52             }
53             public void finmin(String s,int pos){
54                 int cur = 0;
55                 int min = len;
56                 if(pos == len - 1){
57                     results[pos] = 0;
58                 }
59                 else{
60                     for(int i = pos; i < len - 1; i ++){
61                         if(store[pos][i] == true){
62                             cur = 1 + results[i + 1];
63                             if(cur < min){
64                                 min = cur;
65                             }
66                         }
67                     }
68                     if(store[pos][len - 1] == true)
69                         min = 0;
70                     results[pos] = min;
71                 }
72             }
73 }

 

posted on 2013-04-16 12:45  Step-BY-Step  阅读(222)  评论(0)  编辑  收藏  举报

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