LeetCode_Decode Ways-优快云博客

本文链接：https://blog.youkuaiyun.com/leileicaocao/article/details/9183069

本文探讨了如何使用特定映射将字母编码为数字，并详细分析了如何通过递归和动态规划方法计算编码消息的所有可能解码方式。通过实例演示，深入理解编码与解码过程及其背后的数学原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

int DFS(string &s, int step, int len)
{
	if (step == len)
	{
		return 1;
	}
	int count = 0;
	if (s[step] != '0')
	{           
		// 解析一个，只有下一个不为0才可以
		if (!(step + 1 < len && s[step + 1] == '0'))
		{
                
			count += DFS(s, step + 1, len);    
		}        
		// 解析两个，只有s[step] <= 2 
		if (step + 1 < len && s[step - 1] > '0')
		{
			string tmp = s.substr(step - 1, 2);
			if (atoi(tmp.c_str()) <= 26)
				count += DFS(s, step + 2, len);
		}
	}
	return count;
}
int numDecodings(string s) 
{
	if (s == "")
		return 0;
            
	int len = s.length();
	return DFS(s, 0, len);
}

DP：OK

int numDecodings2(string s)
{
    if (s == "")
        return 0;
    int size = s.length();
    int ans = 0;
    int *f = new int[size + 2];
    f[0] = 1;
	f[1] = 1;

    for (int i = 0; i < size; ++i)
    {
        f[i+2] = 0;
        if (s[i] == '0')
        {
            if (i == 0 || s[i-1] == '0' || s[i-1] > '2')
            {
                return 0;
            }
            else
            {
               f[i+2] += f[i];
            }
        }
        else
        {
			// 一个
            f[i+2] += f[i+1];
     
			// 两个			
			if (i > 0 && s[i-1] > '0')
			{
				string tmp = s.substr(i-1, 2);
				if (atoi(tmp.c_str()) <= 26)
				{
					f[i+2] += f[i];				
				}
			}
        }            
    }
    ans = f[size + 1];        
    delete []f;
    return ans;
}