[LintCode] Decode Ways 解码方法

时间:2023-03-09 22:07:28
[LintCode] Decode Ways 解码方法

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.

Example
Given encoded message 12, it could be decoded as AB (1 2) or L (12).
The number of ways decoding 12 is 2.

LeetCode上的原题,请参见我之前的博客Decode Ways

解法一:

class Solution {
public:
/**
* @param s a string, encoded message
* @return an integer, the number of ways decoding
*/
int numDecodings(string& s) {
if (s.empty()) return ;
int n = s.size();
vector<int> dp(n + , );
for (int i = ; i < n + ; ++i) {
if (s[i - ] == '') dp[i] = ;
else dp[i] = dp[i - ];
if (i >= && (s[i - ] == '' || (s[i - ] == '' && s[i - ] <= ''))) {
dp[i] += dp[i -];
}
}
return dp.back();
}
};

解法二:

class Solution {
public:
/**
* @param s a string, encoded message
* @return an integer, the number of ways decoding
*/
int numDecodings(string& s) {
if (s.empty()) return ;
vector<int> dp(s.size() + , );
dp[] = ;
for (int i = ; i < dp.size(); ++i) {
if (s[i - ] >= '' && s[i - ] <= '') dp[i] += dp[i - ];
if (i >= && s.substr(i - , ) <= "" && s.substr(i - , ) >= "") {
dp[i] += dp[i - ];
}
}
return dp.back();
}
};

解法三:

class Solution {
public:
/**
* @param s a string, encoded message
* @return an integer, the number of ways decoding
*/
int numDecodings(string& s) {
if (s.empty() || s.front() == '') return ;
int c1 = , c2 = ;
for (int i = ; i < s.size(); ++i) {
if (s[i] == '') c1 = ;
if (s[i - ] == '' || (s[i - ] == '' && s[i] <= '')) {
c1 = c1 + c2;
c2 = c1 - c2;
} else {
c2 = c1;
}
}
return c1;
}
};