lintcode 解码方法

简单的动态规划

 1 class Solution {
 2 public:
 3     
 4     /*
 5      * @param s: a string,  encoded message
 6      * @return: an integer, the number of ways decoding
 7      */
 8     int numDecodings(string s) {
 9         // write your code here
10         if(s == "" || s[0] == '0') return 0;
11         int dp[110];
12 
13         memset(dp, 0, sizeof(dp));
14         dp[0] = 1;
15         for(int i = 1; i < s.length(); ++i){
16                 if(s[i] == '0'){
17                     if(s[i - 1] != '1' && s[i -1] != '2') return 0;
18                     else {
19                         dp[i] = dp[i] + (i - 2 >= 0 ? dp[i-2] : 1);
20                     }
21                 } else {
22                     if(s[i - 1] == '1'){
23                         dp[i] = dp[i] + (i - 2 >= 0 ? dp[i-2] : 1);
24                         dp[i] = dp[i] + dp[i - 1];
25                     } else if(s[i - 1] == '2'){
26                         if(s[i] <= '6')
27                             dp[i] = dp[i] + (i - 2 >= 0 ? dp[i-2] : 1);
28                         dp[i] = dp[i] + dp[i - 1];
29                     } else {
30                         dp[i] = dp[i] + dp[i - 1];
31                     }
32                 }         
33         }
34 
35         return dp[s.length() - 1];
36     }
37 };

 

posted @ 2017-08-16 18:49  GeniusYang  阅读(234)  评论(0)  编辑  收藏  举报