2014
11-18

# LeetCode-Decode Ways[动态规划]

### Decode Ways

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.

// LeetCode, 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;
}
};


Java代码:

public class Solution {
public int numDecodings(String s) {
if(s.length() == 0) return 0;
if(s.charAt(0) == '0') return 0;
if(s.length() == 1) return s.charAt(0) > '0' ? 1:0;
int dp[] = new int[s.length()+1];
dp[0] = dp[1] = 1;
for(int i=2; i<=s.length(); i++){
dp[i] = dp[i-1];
if(s.charAt(i-1) == '0')
if (s.charAt(i-2) == '1' || s.charAt(i-2) == '2')
dp[i] = dp[i-2];
else return 0;
else if(s.charAt(i-2) == '0'){
dp[i] = dp[i-1];
continue;
}
else if(s.charAt(i-2) < '2' || (s.charAt(i-2) == '2' && s.charAt(i-1) < '7') )
dp[i] += dp[i-2];
}
return dp[s.length()];
}
}