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.
思路：
用动态规划Dynamci Programming来解。一位数时不能为0，两位数不能大于26，其十位上的数也不能为0。用哈希表来存或者用两个变量来存。
State: dp[i]
Function: dp[i] = dp[i - 1] (if i - 1 != 0) + dp[i - 2] (if i - 2 == 1 or i - 2 == 2 and i -1 < = 6)
Initialize: dp[0] = 0, dp[1] = 1
Return: dp[n]
Java 1:
public class Solution {
public int numDecodings(String s) {
if (s.isEmpty() || (s.length() > 1 && s.charAt(0) == ‘0‘)) return 0;
int[] dp = new int[s.length() + 1];
dp[0] = 1;
for (int i = 1; i < dp.length; ++i) {
dp[i] = (s.charAt(i - 1) == ‘0‘) ? 0 : dp[i - 1];
if (i > 1 && (s.charAt(i - 2) == ‘1‘ || (s.charAt(i - 2) == ‘2‘ && s.charAt(i - 1) <= ‘6‘))) {
dp[i] += dp[i - 2];
}
}
return dp[s.length()];
}
}
Java 2:
public class Solution {
public int numDecodings(String s) {
if (s.isEmpty() || (s.length() > 1 && s.charAt(0) == ‘0‘)) return 0;
int prev = 1, prev_prev = 0;
for (int i = 0; i < s.length(); i++) {
int cur = 0;
if (s.charAt(i) != ‘0‘) {
cur = prev;
}
if (i > 0 && (s.charAt(i - 1) == ‘1‘ || (s.charAt(i - 1) == ‘2‘ && s.charAt(i - 1) <= ‘6‘))) {
cur += prev_prev;
}
prev_prev = prev;
prev = cur;
}
return prev;
}
}
Python:
class Solution(object):
def numDecodings(self, s):
if len(s) == 0 or s[0] == ‘0‘:
return 0
prev, prev_prev = 1, 0
for i in xrange(len(s)):
cur = 0
if s[i] != ‘0‘:
cur = prev
if i > 0 and (s[i - 1] == ‘1‘ or (s[i - 1] == ‘2‘ and s[i] <= ‘6‘)):
cur += prev_prev
prev, prev_prev = cur, prev
return prev
