A message containing letters from A-Z
is being encoded to numbers using the following mapping:
'A' -> 1 'B' -> 2 ... 'Z' -> 26
Given a non-empty string containing only digits, determine the total number of ways to decode it.
Example 1:
Input: "12" Output: 2 Explanation: It could be decoded as "AB" (1 2) or "L" (12).
Example 2:
Input: "226" Output: 3 Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
思路:
动态规划:
dp[i] = dp[i-1]+dp[i-1] if s[i-1], s[i-2] are valid
dp[i] = dp[i-1] if only dp[i-1] valid
代码如下:
int numDecodings(string s) {
if (s.size() == 0) return 0;
vector<int> dp(s.size(), 0);
for (int i = 0; i < s.size(); i++){
if (s[i] == '0'){
bool situation1 = i>=2 && s[i - 1] > '0' && s[i-1] <= '2';
bool situation2 = i == 1 && s[i - 1] > '0' && s[i - 1] <= '2';
if (situation1) dp[i] = dp[i - 2];
else if (situation2) dp[i] = dp[i - 1];
else dp[i] = 0;
}
else{
if (i > 1){
bool situation1 = s[i - 1] == '1';
bool situation2 = s[i - 1] == '2' && s[i] <= '6';
if (situation1 || situation2) dp[i] = dp[i - 1] + dp[i - 2];
else dp[i] = dp[i - 1];
}
else if (i == 1){
bool situation1 = s[i - 1] == '1';
bool situation2 = s[i - 1] == '2' && s[i] <= '6';
if (situation1 || situation2) dp[i] = dp[i - 1] + 1;
else dp[i] = dp[i - 1];
}
else{
dp[i] = 1;
}
}
}
return dp[s.size() - 1];
}
纪念贴图: