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.
给定数字串,判断由字母组成的情况的数量。
简单DP,当前位为0,不可能由前一位转移,若前一位为和当前为组合小于26,则可以用之前两位的状态转移。
class Solution {
public:
int numDecodings(string s) {
if(s.empty()) return 0;
vector<int> dp(s.size()+1,0);
dp[0]=1;
if(s[0]!='0') dp[1]=1;
for(int i=1;i<s.size();i++)
{
if(s[i]!='0') dp[i+1]+=dp[i];
if((s[i]>='0' && s[i]<='9' && s[i-1]=='1') || (s[i-1]=='2' && s[i]>='0' && s[i]<='6')) dp[i+1]+=dp[i-1];
}
return dp[s.size()];
}
};