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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.
实例:10应返回1, 01应返回0, 30应返回0。
题目总感觉在涉及0的时候有分歧,通过这几个实例就能体会到。
分析:需要注意的是,如果序列中有不能匹配的0,那么解码方法是0,比如序列012 、100(第二个0可以和1组成10,第三个0不能匹配)。递归的解法很容易,但是大集合会超时。转换成动态规划的方法,假设dp[i]表示序列s[0...i-1]的解码数目,动态规划方程如下:
代码如下:
1 class Solution { 2 public: 3 int numDecodings(string s) { 4 // IMPORTANT: Please reset any member data you declared, as 5 // the same Solution instance will be reused for each test case. 6 //注意处理字符串中字符为0的情况 7 int len = s.size(); 8 if(len == 0)return 0; 9 int dp[len+1];//dp[i]表示s[0...i-1]的解码方法数目 10 dp[0] = 1; 11 if(s[0] != ‘0‘)dp[1] = 1; 12 else dp[1] = 0; 13 for(int i = 2; i <= len; i++) 14 { 15 if(s[i-1] != ‘0‘) 16 dp[i] = dp[i-1]; 17 else dp[i] = 0; 18 if(s[i-2] == ‘1‘ || (s[i-2] == ‘2‘ && s[i-1] <= ‘6‘)) 19 dp[i] += dp[i-2]; 20 } 21 return dp[len]; 22 } 23 };
http://www.cnblogs.com/TenosDoIt/p/3451920.html
标签:style blog http color io os ar strong for
原文地址:http://www.cnblogs.com/forcheryl/p/3991738.html