Leetcode: Word Ladder

本文介绍了一个算法问题,即寻找两个单词之间的最短转换路径。利用宽度优先搜索(BFS)策略,从起始单词开始,每次只改变一个字母,并确保每个中间状态都在给定的字典中。最终达到目标单词时返回转换路径的长度。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that:
Only one letter can be changed at a time
Each intermediate word must exist in the dictionary

For example,

Given:
start = "hit"
end = "cog"
dict = ["hot","dot","dog","lot","log"]

As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

Note:

  • Return 0 if there is no such transformation sequence.
  • All words have the same length.
  • All words contain only lowercase alphabetic characters.
BFS: 

int ladderLength(string start, string end, unordered_set<string> &dict) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
		if(start.size() != end.size()) return 0;
		if(start.empty() || end.empty())return 0;
		
		queue<string> path;
		path.push(start);
		int level = 1;
		int count = 1;
		dict.erase(start);
		while(dict.size() > 0 && !path.empty())
		{
			string curword = path.front();
			path.pop();count--;
			for(int i = 0; i < curword.size(); i++)
			{
				string tmp = curword;
				for(char j='a'; j<='z'; j++)
				{
					if(tmp[i]==j)continue;
					tmp[i] = j;
					if(tmp==end)return level+1;
					if(dict.find(tmp) != dict.end()) path.push(tmp);
					dict.erase(tmp);
				}
			}
			if(count==0)
			{
				count = path.size();
				level++;
			}
		}
		return 0;
    }





评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值