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.
思路：
典型的动规，用n[i]表示从i到末尾的表示法，然后对于每一个i，判断s[i]和s[i]+s[j]能否解码，再分别加上n[i+1]和n[i+2]即可。
代码：
1 int numDecodings(string s) { 2 // IMPORTANT: Please reset any member data you declared, as 3 // the same Solution instance will be reused for each test case. 4 int l = s.length(); 5 if(l == 0) 6 return 0; 7 if(l == 1) 8 return s[0] > '0'; 9 int ways[l]; 10 int i; 11 ways[l-1] = s[l-1] > '0'; 12 if(s[l-2] == '0') 13 ways[l-2] = 0; 14 else{ 15 if((s[l-2]-'0')*10+s[l-1]-'0' <= 26) 16 ways[l-2] = ways[l-1] + 1; 17 else 18 ways[l-2] = ways[l-1]; 19 } 20 for(i = l-3; i >= 0; i--){ 21 if(s[i] == '0'){ 22 ways[i] = 0; 23 continue; 24 } 25 int tmp = (s[i]-'0')*10+s[i+1]-'0'; 26 if(tmp <= 26) 27 ways[i] = ways[i+1] + ways[i+2]; 28 else 29 ways[i] = ways[i+1]; 30 } 31 return ways[0]; 32 }
第二遍
1 int numDecodings(string s) { 2 // IMPORTANT: Please reset any member data you declared, as 3 // the same Solution instance will be reused for each test case. 4 int l = s.length(); 5 if(l == 0) 6 return 0; 7 if(l == 1) 8 return s[0]>'0'; 9 int result[l+1]; 10 result[l-1] = s[l-1]>'0'; 11 result[l] = 1; 12 for(int i = l-2; i >= 0; i--){ 13 if(s[i] == '0') 14 result[i] = 0; 15 else{ 16 int tmp = 10*(s[i]-'0')+s[i+1]-'0'; 17 if(tmp <= 26) 18 result[i] = result[i+1]+result[i+2]; 19 else 20 result[i] = result[i+1]; 21 } 22 } 23 return result[0]; 24 }