題目:
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.
題解:
動態規划來做。
設置動態數組dp[n+1]。dp[i]表示從1~i的decode ways的個數。
當給的code只有一位數時,判斷是不是valid(A~Z),是的話就dp[1] = 1 不是的話就是dp[1] = 0
因為像給的例子12可以有兩種可能的解析方法,所以計算dp[i]的時候要判斷兩種可能性,再累加。
代碼如下:
1
public
int numDecodings(String s) {
2 if (s.length()==0||s== null||s=="0")
3 return 0;
4
5 int[] dp = new int[s.length()+1];
6 dp[0] = 1;
7
8 if (isValid(s.substring(0,1)))
9 dp[1] = 1;
10 else
11 dp[1] = 0;
12
13 for( int i=2; i<=s.length();i++){
14 if (isValid(s.substring(i-1,i)))
15 dp[i] += dp[i-1];
16 if (isValid(s.substring(i-2,i)))
17 dp[i] += dp[i-2];
18 }
19 return dp[s.length()];
20 }
21
22 public boolean isValid(String s){
23 if (s.charAt(0)=='0')
24 return false;
25 int code = Integer.parseInt(s);
26 return code>=1 && code<=26;
27 }
2 if (s.length()==0||s== null||s=="0")
3 return 0;
4
5 int[] dp = new int[s.length()+1];
6 dp[0] = 1;
7
8 if (isValid(s.substring(0,1)))
9 dp[1] = 1;
10 else
11 dp[1] = 0;
12
13 for( int i=2; i<=s.length();i++){
14 if (isValid(s.substring(i-1,i)))
15 dp[i] += dp[i-1];
16 if (isValid(s.substring(i-2,i)))
17 dp[i] += dp[i-2];
18 }
19 return dp[s.length()];
20 }
21
22 public boolean isValid(String s){
23 if (s.charAt(0)=='0')
24 return false;
25 int code = Integer.parseInt(s);
26 return code>=1 && code<=26;
27 }
Reference:
http://blog.csdn.net/u011095253/article/details/9248109
