Leetcode Decode Ways 91

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最新推荐文章于 2025-08-20 09:37:41 发布

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分类专栏：算法练习文章标签： leetcode

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本文介绍了一种将字母编码为数字的映射方法，并详细解释了如何计算给定数字编码消息的所有可能解码方式的数量。通过动态规划算法实现了解码计数的功能。

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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.

题目链接

考虑对于s[i]
若本身>=0 && <=9 则 dp[i]=dp[i-1]
若s[i-1]*10+s[i] >=10 && <=26 则dp[i]=dp[i-1]+dp[i-2]

class Solution {
public:
    int numDecodings(string s) {
        int dp[10000];
        int len=s.length();

        for(int i=0;i<10000;i++)
            dp[i]=0;
        if(judge1((int)(s[0]-'0'))){
            dp[0]=1;
        }
        else
            return 0;
        if(len==1) return dp[0];

        int digit=((int)(s[1]-'0'));
        int temp=((int)(s[0]-'0'))*10+((int)(s[1]-'0'));
        if(judge1(digit)&&judge2(temp)){
            dp[1]=2;
        }
        else if(judge1(digit)&&(!judge2(temp))){
            dp[1]=1;
        }
        else if((!judge1(digit))&&judge2(temp)){
            dp[1]=1;
        }
        else return 0;

        for(int i=2;i<len;i++){
            int digit=((int)(s[i]-'0'));
            int temp=((int)(s[i-1]-'0'))*10+((int)(s[i]-'0'));
            if(judge1(digit)&&judge2(temp)){
                dp[i]=dp[i-1]+dp[i-2];
            }
            else if(judge1(digit)&&(!judge2(temp))){
                dp[i]=dp[i-1];
            }
            else if((!judge1(digit))&&judge2(temp)){
                dp[i]=dp[i-2];
            }
        }

        return dp[len-1];
    }

    bool judge1(int one){
        if(one>=1&&one<=9) return true;
        return false;
    }

    bool judge2(int two){
        if(two>=10&&two<=26) return true;
        return false;
    }
};