Decode Ways
描述
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.
分析
跟第???节Climbing Stairs很类似,不过多加一些判断逻辑。
代码
// Decode Ways
// 动规,时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
int numDecodings(const string &s) {
if (s.empty() || s[0] == '0') return 0;
int prev = 0;
int cur = 1;
// 长度为n的字符串,有 n+1个阶梯
for (size_t i = 1; i <= s.size(); ++i) {
if (s[i-1] == '0') cur = 0;
if (i < 2 || !(s[i - 2] == '1' ||
(s[i - 2] == '2' && s[i - 1] <= '6')))
prev = 0;
int tmp = cur;
cur = prev + cur;
prev = tmp;
}
return cur;
}
};