题目描述:
A message containing letters from A-Z
is being encoded to numbers using the following mapping way:
'A' -> 1 'B' -> 2 ... 'Z' -> 26
Beyond that, now the encoded string can also contain the character '*', which can be treated as one of the numbers from 1 to 9.
Given the encoded message containing digits and the character '*', return the total number of ways to decode it.
Also, since the answer may be very large, you should return the output mod 109 + 7.
Example 1:
Input: "*" Output: 9 Explanation: The encoded message can be decoded to the string: "A", "B", "C", "D", "E", "F", "G", "H", "I".
Example 2:
Input: "1*" Output: 9 + 9 = 18
Note:
题意大概是这么个意思:给一个只包含*和1-9的字符串,其中星号*可以任意替换为1-9。如果用A-Z表示1-26,那么找到给定字符串可以由多少种只由A-Z的字符串转换得到。
思路一:
首先考虑:一位数字可以表示一个字母,比如1-A,两位字符也可以表示一个字母比如26-Z。那么根据是1个或者2个字符表示1个字母很快可以想到一个DFS的思路。
对于每个大于2个字符的字符串S[n],我们可以把它分解成两种情况:
1. 前面n-2个字符的子串 和 最后两个字符的子串。
2. 前面n-1个字符的子串 和 最后一个字符的子串。
如果用nums()表示A-Z字符串可以匹配的个数,我们可以得到
nums(S[1..n]) = nums(S[1...n-2] ) * nums(S[n-1,n]) + nums(S[1...n-1] ) * nums(S[n]) .
DFS 终止条件为:当要找的字符串长度小于3的时候,我们通过单独的分析一一列举出来。
思路二:
当一个题目可以被DFS搞定,接下来我们就得找找有没有对应的DP算法了。通过上文我们知道。S[1..n] 总是可以由 S[1...n-1] + S[1....n-2]推导出来。无非是多了个系数:中间再加上长度为1和2的字符串单独分析过程。Bingo,这不就是变形版的fibonacci数列问题吗!!
话不多说,直接上代码,参考lc大牛:
class Solution {
public:
int numDecodings(string s) {
int n = s.size(), p = 1000000007;
// f2 is the answer to sub string ending at position i; Initially i = 0.
long f1 = 1, f2 = helper(s.substr(0,1));
// DP to get f2 for sub string ending at position n-1;
for (int i = 1; i < n; i++) {
long f3 = (f2*helper(s.substr(i, 1)))+(f1*helper(s.substr(i-1, 2)));
f1 = f2;
f2 = f3%p;
}
return f2;
}
private:
int helper(string s) {
if (s.size() == 1) {
if (s[0] == '*') return 9;
return s[0] == '0'? 0:1;
}
// 11-26, except 20 because '*' is 1-9
if (s == "**")
return 15;
else if (s[1] =='*') {
if (s[0] =='1') return 9;
return s[0] == '2'? 6:0;
}
else if (s[0] == '*')
return s[1] <= '6'? 2:1;
else
// if two digits, it has to be in [10 26]; no leading 0
return stoi(s) >= 10 && stoi(s) <= 26? 1:0;
}
};