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.
动态规划的思路:
如果当前位置n的字符串若不为0,那么包含前n-1个字符串组成的解码个数;
若n-1位的字符串不为0,且n-1和n位可组成小于等于26的2位数,包含前n-2个字符串组成的解码个数。
class Solution {
public:
int numDecodings(string s) {
int n = s.size();
if(n==0) return 0;
else if(n == 1)
return (s[0]!='0')?1:0;
vector<int> f(n);
f[0] = (s[0]!='0')?1:0;
f[1] = (s[1]!='0' && s[0]!='0'?1:0) + ((s[0]-'0')*10+(s[1]-'0')<=26 && (s[0]!='0')? 1:0);
for(int i = 2; i < n; i++)
{
f[i] = ((s[i]!='0')?f[i-1]:0) + ((s[i-1]-'0')*10+(s[i]-'0')<=26 && (s[i-1]!='0')? f[i-2]:0);
}
return f[n-1];
}
};