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).
Solutions
class Solution {
public:
int numDecodings(string s) {
vector<int> n(s.size() + 1, 0);
if (s[0] == '0')
return 0;
// s[0:i]: first i elements in s (not including s[i])
// n[i]: number of ways to decode s[0:i]
n[0] = n[1] = 1;
for (int i = 1; i < s.size(); ++i) {
if (s[i] == '0') {
if (s[i - 1] == '1' || s[i - 1] == '2') {
n[i + 1] = n[i - 1];
} else {
return 0;
}
} else if (s[i] < '7') {
if (s[i - 1] == '1' || s[i - 1] == '2') {
n[i + 1] = n[i] + n[i - 1];
} else {
n[i + 1] = n[i];
}
} else {
if (s[i - 1] == '1') {
n[i + 1] = n[i] + n[i - 1];
} else {
n[i + 1] = n[i];
}
}
}
return n[s.size()];
}
};
More concise solution
class Solution {
public:
int numDecodings(string s) {
if (s[0] == '0')
return 0;
vector<int> dp(s.size() + 1);
dp[0] = dp[1] = 1;
for (int i = 1; i < s.size(); ++i) {
// temp: number of ways to decode s[0:i+1]
// such that s[i] is alone.
int temp = s[i] == '0' ? dp[i] : 0;
if (s[i - 1] == '1' || s[i - 1] == '2' && s[i] < '7') {
// possible to group s[i-1] and s[i]
dp[i + 1] = temp + dp[i - 1];
} else {
// not possible to group s[i-1] and s[i]
dp[i + 1] = temp;
}
}
return dp[s.size()];
}
};
