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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.
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题意：A->1 B->2 ...Z->26;给出一个数字字符串，求可能的驿码方式
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public
class
Solution {
public
int
numDecodings(String s) {
int
length=s.length();
if
(s==
null
|| length==
0
)
return
0
;
int
[] dp=
new
int
[length+
1
];
dp[
0
]=
1
;
dp[
1
]=s.charAt(
0
)==
'0'
?
0
:
1
;
for
(
int
i=
2
;i<=length;i++){
int
first=Integer.valueOf(s.substring(i-
1
,i));
int
second=Integer.valueOf(s.substring(i-
2
,i));
if
(first>
0
&&first<
10
){
dp[i]+=dp[i-
1
];
}
if
(second>=
10
&& second<=
26
){
dp[i]+=dp[i-
2
];
}
}
return
dp[length];
}
}
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PS:看大神说可以用dp...第一次用。。其实有一小点还不明白，dp[0]=1是为什么呢
本文转自 努力的C 51CTO博客，原文链接:http://blog.51cto.com/fulin0532/1902575
