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  • Decode Ways——动态规划

    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.

    这题使用动态规划来做,但是刚开始的时候一直超时,我以为是其他地方的原因,但是怎么改都有问题(代码如下)后来,我发现我这样做压根就不是动态规划啊,就是普通的递归,怪不得会超时·················动态规划与普通的递归的区别就是在于不要重复计算啊!!!!!动态规划得到初始的0,1,然后由0,1计算得到后面的数·······

    class Solution {
    public:
        int numDecodings(string s) {
            int len=s.size();
            if(len==0||s[0]=='0')
                return 0;
            int num;
            int res;
            if(len==1)
                num =s[0]-'0';
            else if(len==2)
                num=10*(s[0]-'0')+s[1]-'0';
            if(num==0)
                return 0;
            else if(num<=10)
                return 1;
            else if(num>10&&num<=26)
                return 2;
            else if(num>26&&num<=99)
                return 1;
            res=max(numDecodings(s.substr(0,1))+numDecodings(s.substr(1,len)),
                    numDecodings(s.substr(0,2))+numDecodings(s.substr(2,len)));
            return res;
        }
    };

     正确的做法应该是这样的:

    class Solution {
    public:
        int numDecodings(string s) {
            int len=s.size();
            if(len==0)
                return 0;
            else if(len==1)
                return s[0]!='0'?1:0;
            else if(len==2)
                return (s[0]!='0'&&s[1]!='0'?1:0)+(s[0]!='0'&&((10*(s[0]-'0')+s[1]-'0')<=26)?1:0);
                
                
            int* flag=new int[len];
            flag[0]=s[0]!='0'?1:0;
            flag[1]=(s[0]!='0'&&s[1]!='0'?1:0)+(s[0]!='0'&&((10*(s[0]-'0')+s[1]-'0')<=26)?1:0);
            for(int i=2;i<len;i++)
            {
                flag[i]=0;
                if(s[i]!='0')
                {
                    flag[i]+=flag[i-1];
                }
                if(s[i-1]!='0'&&(10*(s[i-1]-'0')+s[i]-'0'<=26))
                {
                    flag[i]+=flag[i-2];
                }
            }
            int res=flag[len-1];
            delete []flag;
            return res;
        }
    };
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  • 原文地址:https://www.cnblogs.com/qiaozhoulin/p/4605732.html
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