LeetCode 91: Decode Ways-优快云博客

本文介绍了一种用于解码数字字符串的算法，该字符串由字母A到Z编码为1到26的数字。通过动态规划方法计算给定编码消息的所有可能解码方式的数量，并详细解析了实现过程中的特殊案例。

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Difficulty: 3

Frequency: 4

Problem:

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.

Solution:

class Solution {
public:
    int numDecodings(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (s.size()==0||s[0]=='0')// Subtlety 1
            return 0;
            
        vector<int> decode_method(s.size()+1);
        decode_method[0] = 1;
        decode_method[1] = 1;
        
        for (int i = 1; i<s.size(); i++)
        {
            if (s[i]=='0')
            {
                if (s[i-1]=='0'||s[i-1]>'2')// Subtlety 2
                    return 0;
                    
                decode_method[i+1] = decode_method[i-1];
            }
            else
            {
                switch(s[i-1]){// Subtlety 3
                    case '1':
                        decode_method[i+1] = decode_method[i] + decode_method[i-1];
                        break;
                    case '2':
                        if (s[i]<='6'&&s[i]>='0')
                            decode_method[i+1] = decode_method[i] + decode_method[i-1];
                        else
                            decode_method[i+1] = decode_method[i];
                        break;
                    default:
                        decode_method[i+1] = decode_method[i];
                        break;
                }
            }
        }
        return decode_method[s.size()];
    }
};

Notes:

Kinda like Fibonacci numbers. But there are lots of special cases should be pay attention to.