Leetcode——91. Decode Ways
2017-03-10 23:53
316 查看
#
https://leetcode.com/problems/decode-ways/?tab=SolutionsA 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.
#
动态规划问题:递推公式为DP问题:dp[i+2]=s[i,i+1]<=26?2*dp[i]:dp[i]
dp[i]表示到i位置,可能的解码顺序。
可是这一题需要考虑300,00,这种有0出现的情况,这些边边角角真TM烦人!
本来十来行代码,现在为了检查意外情况一下子变成几十行了!
class Solution {//DP问题:dp[i+2]=(s[i,i+1]<=26?2*dp[i]:dp[i]) public: int numDecodings(string s) { if(s.length()==0) return 0; int *dp=new int[s.length()]; int *num=new int[s.length()]; for(int i=0;i<s.length();i++) { num[i]=s[i]-'0'; } //合理性检查,排除30,00这种数的出现 if(num[0]==0) return 0; else for(int i=1;i<s.length();i++) { if((num[i]==0&&num[i-1]>=3)||(num[i]==0&&num[i-1]==0)) return 0; else continue; } dp[0]=1; if(s.length()>=2) { int temp=num[0]*10+num[1]; if(temp<=26&&num[1]!=0) dp[1]=2; else dp[1]=1; } for(int i=2;i<s.length();i++) { if(num[i-1]==0)//前面的判断已经保证不会出现00 dp[i]=dp[i-1]; else if(num[i]==0) { if(num[i-1]<3)//出现20这种 dp[i]=dp[i-2]; else return 0; } else { int temp=num[i-1]*10+num[i]; if(temp<=26) dp[i]=dp[i-1]+dp[i-2]; else dp[i]=dp[i-1]; } } return dp[s.length()-1]; } };
相关文章推荐
- [leetcode] 91. Decode Ways
- [leetcode] 91. Decode Ways 解题报告
- [leetcode]91. Decode Ways
- LeetCode-91. Decode Ways
- Leetcode 91. Decode Ways 解码方法(动态规划,字符串处理)
- [Leetcode] 91. Decode Ways
- Leetcode 91. Decode Ways
- LeetCode 91. Decode Ways(解码方法)
- [LeetCode] 91. Decode Ways
- [leetcode]91. Decode Ways(Java)
- Leetcode 91. Decode Ways 解码方法(动态规划,字符串处理)
- [LeetCode] 91. Decode Ways 解码方法
- leetcode-91. Decode Ways
- Leetcode 91. Decode Ways JAVA语言
- Leetcode-91. Decode Ways
- Leetcode 91. Decode Ways
- [LeetCode] 91. Decode Ways
- LeetCode 91. Decode Ways(解码方法)
- LeetCode 91. Decode Ways
- 个人记录-LeetCode 91. Decode Ways