91. Decode Ways
91. Decode Ways
1. 题目
A message containing letters from A-Z
is being encoded to numbers using the following mapping:
'A' -> 1
'B' -> 2
...
'Z' -> 26
Given a non-empty string containing only digits, determine the total number of ways to decode it.
Example 1:
Input: "12"
Output: 2
Explanation: It could be decoded as "AB" (1 2) or "L" (12).
Example 2:
Input: "226"
Output: 3
Explanation: It could be decoded as "BZ" (2 26), "VF" (22 6), or "BBF" (2 2 6).
2. 思路
本题求解的是在给定的条件下(既前文中A->1,B->2)给到一个数字,比如226 有多少种解释方法。这题可以使用动态规划的方法来求解。假设dp[i]为以i为结尾的字符串解码方式数量的总和。那么dp[i]的递推规则是:
- 如果s[i-1] != "0",则dp[i] += dp[ i - 1 ] ,代表了一次解一位,则剩下n-1个字符需要解
- 如果一次解2位,则要求是s[i-2:i] <"27" and s[i-2:i] >'09',dp[i] += dp[i-2]
3.实现
class Solution(object):
def numDecodings(self, s):
"""
:type s: str
:rtype: int
"""
if s == '':
return 0
dp = [0 for x in range(len(s)+1)]
dp[0] = 1
for i in range(1, len(s)+1):
if s[i-1] != "0":
dp[i] += dp[i-1]
if i != 1 and s[i-2:i] < "27" and s[i-2:i] > "09":
dp[i] += dp[i-2]
return dp[-1]
编程光~