Leetcode 91. Decode Ways

91. 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.

/*
题意：字母表A-Z对应数字1-26，给定字符串S，问有多少种解码方式
思路：设状态dp[i]表示字符串前i个字符能表示的解码个数。那么它依赖于dp[i-1]和dp[i-2]
具体的依赖关系如下：
（1）s[i] == '0'，如果满足s[i-1] == '1' || s[i-1] == '2' 则dp[i] = dp[i-2]，否则返回0
（2）不满足条件1，但是满足s[i-1] == '1' || (s[i-1] == '2' && s[i] <= '6')则
dp[i] = dp[i-1] + dp[i-2]
（3）不满足前两个条件，则dp[i] = dp[i-1]
*/
class Solution {
public:
int numDecodings(string s) {
int len = s.length();
if(!len) return 0;
vector<int> dp(len + 1, 0);
dp[0] = 1;
for(int i = 1; i <= len; i++){
if(s[i-1] == '0'){
if(i > 1 && (s[i-2] == '1' || s[i-2] == '2'))
dp[i] = dp[i-2];
else
return 0;
}
else if(i > 1 && (s[i-2] == '1' || (s[i-2] == '2' && s[i-1] <= '6')))
dp[i] = dp[i-1] + dp[i-2];
else
dp[i] = dp[i-1];
}
return dp[len];
}
};