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leetcode--Decode Ways

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.

For 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        int num = 0;
 4         if(s.length() > 0){
 5             int[] decodeways  = new int[s.length()];
 6             if(Integer.parseInt(s.substring(0, 1)) == 0)
 7                 return 0;
 8             else
 9                 decodeways[0] = 1;            
10             if(s.length()> 1){
11                 if(Integer.parseInt(s.substring(1, 2)) != 0){
12                     if(Integer.parseInt(s.substring(0, 2)) > 26)
13                         decodeways[1] = 1;
14                     else{
15                         if(Integer.parseInt(s.substring(0, 1)) == 0)
16                             return 0;
17                         else
18                             decodeways[1] = 2;                            
19                     }                
20                 }                
21                 else{
22                     if(Integer.parseInt(s.substring(0, 1)) > 2)
23                         return 0;
24                     else
25                         decodeways[1] = 1;                        
26                 }
27                 for(int i = 2; i < s.length(); ++i){
28                     if(Integer.parseInt(s.substring(i, i + 1)) != 0){
29                         if(Integer.parseInt(s.substring(i - 1, i + 1)) > 26 || 
30                             Integer.parseInt(s.substring(i - 1, i)) == 0)
31                             decodeways[i] = decodeways[i - 1];
32                         else
33                             decodeways[i] = decodeways[i - 1] + decodeways[i - 2];
34                     }
35                     else{
36                         if(Integer.parseInt(s.substring(i - 1, i)) == 0 || 
37                             Integer.parseInt(s.substring(i - 1, i)) > 2){
38                             num = 0;
39                             break;
40                         }
41                         else
42                             decodeways[i] = decodeways[i - 2];
43                     }                    
44                 }
45             }
46             num = decodeways[s.length() - 1];
47         }
48         return num;    
49     }
50 }

code 2, the same algorithm as above

 1 public class Solution {
 2     public int numDecodings(String s) {
 3       int num = 0;
 4         if(s.length() == 0 || s.charAt(0) == '0')
 5             return num;
 6           else{
 7             int[] ways = new int[s.length() + 1];
 8             ways[0] = ways[1] = 1;
 9             for(int i = 2; i < s.length() + 1; ++i){
10                 int cur = Integer.parseInt(s.substring(i - 1, i));
11                 int pre = Integer.parseInt(s.substring(i - 2, i - 1));
12                 if(cur != 0){
13                     if(pre != 0){
14                         ways[i] = (Integer.parseInt(s.substring(i - 2, i)) < 27) ? ways[i - 1] + ways[i - 2] : ways[i - 1];
15                     }
16                     else
17                         ways[i] = ways[i - 2];                        
18                 }
19                 else{
20                     if(pre > 2 || pre == 0)
21                         return num;
22                     else
23                         ways[i] = ways[i - 2];
24                 }    
25             }
26             num = ways[s.length()];    
27         }
28         return num;
29     }
30 }

The third code:

public class Solution {
    public int numDecodings(String s) {
       int result = 0;
		int length = s.length();
		if(length > 0){
			int[] ways = new int[length + 1];
			ways[length] = 1;
			ways[length - 1] = (s.charAt(length - 1) - '0' == 0) ? 0 : 1;
			for(int i = length - 2; i > -1; --i){
				if(s.charAt(i)-'0' == 0)
					ways[i] = 0;
				else if((s.charAt(i) - '0' == 1)){
					if(s.charAt(i + 1) - '0' == 0)
						ways[i] = ways[i + 2];
					else
						ways[i] = ways[i + 2] + ways[i + 1];
				}
				else if(s.charAt(i) -'0' == 2){
					if(s.charAt(i + 1) - '0' == 0)
						ways[i] = ways[i + 2];
					else if(s.charAt(i + 1) - '0' != 0 && s.charAt(i + 1) - '0' < 7)
						ways[i] = ways[i + 1] + ways[i + 2];
					else
						ways[i] = ways[i + 1];					
				}
				else{
					if(s.charAt(i + 1) - '0' == 0)
						ways[i] = 0;
					else
						ways[i] = ways[i + 1];
				}
			}
			result = ways[0];
		}
		return result;
    }
}

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原文地址：https://www.cnblogs.com/averillzheng/p/3536707.html