Leetcode Decode Ways
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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.
看到题目,一下想到的思路就是使用DFS,然后可以使用向量可以存储所有的结果。但后面仔细看了看题目,不需要列出所有的结果,只需要输出数字串转换的结果个数,所以又到了动态规划大法登场的时候。
这个动态方程比较复杂:分为两个字符和一个字符的情况,假设当前索引为i,字符串为s,n为字符串的长度,valid1表示一个字符的情况判断函数,valid2标识两个字符的判断情况,dp[n+1]为动态数组,dp数组的初始化需要注意
1.当valid1()==true && valid2() == true; dp[i] = dp[i-1] +dp[i-2]
2.当valid1()==true && valid2() == false,dp[i] = dp[i-1];
3.当valid1() == false && valid2() == true,dp[i] = dp[i-2];
4.当valid1() == false && valid2() == false.return 0;
代码如下:
class Solution {public: bool valid(char a,char b) { return a == '1'||(a=='2'&&b<='6'); } bool valid(char a) { return a!='0'; } int numDecodings(string s) { int len = s.length(); if(len == 0) return 0; int dp[len+1]; memset(dp,0,sizeof(int)*(len+1)); dp[0] = 1; if(s[0] == '0') return 0; else dp[1] =1; for(int i=1;i<len;i++) { if(valid(s[i-1],s[i]) && valid(s[i])) dp[i+1] = dp[i]+dp[i-1]; if(!valid(s[i-1],s[i]) && valid(s[i])) dp[i+1] = dp[i]; if(valid(s[i-1],s[i]) && !valid(s[i])) dp[i+1] = dp[i-1]; if(!valid(s[i-1],s[i]) && !valid(s[i])) return 0; } return dp[len]; }};
发现前面居然有更快的就看了一下更快的代码,发现思路是一致的,但是它的写法更简洁,也缺少了函数的调用,代码如下:
class Solution {public: int numDecodings(string s) { if (!s.size() || s.front() == '0') return 0; int r1 = 1, r2 = 1; for (int i = 1; i < s.size(); i++) { if (s[i] == '0') r1 = 0; if (s[i - 1] == '1' || s[i - 1] == '2' && s[i] <= '6') { int tmp = r1; r1 = r2 + r1; r2 = tmp; } else { r2 = r1; } } return r1; }};
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