[LintCode] Decode Ways

阿新 • • 發佈：2017-06-01

ron ane con eterm number panel pre str using

A message containing letters from A-Z is being encoded to numbers using the following mapping:

‘A‘ -> 1
‘B‘ -> 2
...
‘Z‘ -> 26

Given an encoded message containing digits, determine the total number of ways to decode it.

Example

Given encoded message 12, it could be decoded as AB (1 2) or L

(12).
The number of ways decoding 12 is 2.

 1 public class Solution {
 2     public int numDecodings(String s) {
 3         if(s == null || s.length() == 0){
 4             return 0;
 5         }
 6         return dfs(s, s.length() - 1);   
 7     }
 8     private int dfs(String s, int end){
 9         // 
base case of length 1
10         if(end == 0){
11             if(s.charAt(end) != ‘0‘){
12                 return 1;
13             }
14             else{
15                 return 0;
16             }
17         }
18         //base case of length 2
19         if(end == 1){
20             int ways = 0;
21             if 
(s.charAt(end - 1) == ‘0‘){
22                 return 0;
23             }
24             if(s.charAt(end) != ‘0‘){
25                 ways++;
26             }
27             int num = (s.charAt(end - 1) - ‘0‘) * 10 + (s.charAt(end) - ‘0‘);
28             if(num <= 26){
29                 ways++;
30             }
31             return ways;
32         }
33         int ways = 0;
34         if(s.charAt(end) != ‘0‘){
35             ways += dfs(s, end - 1);
36         }
37         int num = (s.charAt(end - 1) - ‘0‘) * 10 + (s.charAt(end) - ‘0‘);
38         if(s.charAt(end - 1) != ‘0‘ && num <= 26){
39             ways += dfs(s, end - 2);
40         }
41         return ways;
42     }
43 }

 1 public class Solution {
 2     public int numDecodings(String s) {
 3         if(s == null || s.length() == 0){
 4             return 0;
 5         }
 6         int n = s.length();
 7         int[] states = new int[n + 1];
 8         states[0] = 1;
 9         states[1] = (s.charAt(0) == ‘0‘ ? 0 : 1);
10         for(int i = 2; i <= n; i++){
11             states[i] = 0;
12         }
13         for(int i = 2; i <= n; i++){
14             int num = (s.charAt(i - 2) - ‘0‘) * 10 + (s.charAt(i - 1) - ‘0‘);
15             if(s.charAt(i - 1) != ‘0‘){
16                 states[i] += states[i - 1];
17             }
18             if(s.charAt(i - 2) != ‘0‘ && num <= 26){
19                 states[i] += states[i - 2];
20             }
21             if(states[i] == 0){
22                 return 0;
23             }
24         }
25         return states[n];
26     }
27 }

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