[leetcode] 127. Word Ladder 解题报告

题目链接: https://leetcode.com/problems/word-ladder/

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord toendWord, such that:

1. Only one letter can be changed at a time
2. Each intermediate word must exist in the word list

For example,

Given:
beginWord = "hit"
endWord = "cog"
wordList = ["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.

思路: 这个就是图论算法中的单源最短路, 求单源最短路比较通用的算法是BFS和Dijkstra, 其区别是BFS不能用于带权重的图中, 而后者可以, 可以看出在本题中两个字符串之间是无权重的, 也就是如果连通就是1, 不联通就是无穷. BFS和Dijkstra的区别是前者的时间复杂度是O(n), 后者最多优化到O(m log n), 所以如果条件成立一般选择BFS要更好.

其思路就是先把起点加到队列中, 然后每次将字典中与队首距离为1的字符串加进队列, 直到最后出队列的是终点字符串, 为确保终点字符串存在, 我们可以先在字典中加进去终点字符串.

而在本题中在寻找与一个字符串相距为1的的字典中另一个字符串时如果一个个遍历字典消耗时间比较多, 每次时间复杂度是O(n). 在单个字符串不是很长的情况下, 一个个查看改变一个字符然后在字典中查看是否存在效率要更高些, 其时间复杂度是O(k log n), 其中k为单个字符串长度, n为字典长度.

代码如下:

class Solution {
public:
    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordList) {
        wordList.insert(endWord);
        queue<pair<string, int>> que;
        que.push(make_pair(beginWord, 1));
        wordList.erase(wordList.find(beginWord));
        while(!que.empty())
        {
            auto val = que.front();
            que.pop();
            if(val.first == endWord) return val.second;
            for(int i =0; i< val.first.size(); i++)
            {
                string str = val.first;
                for(int j = 0; j < 26; j++)
                {
                    str[i] = 'a'+j;
                    if(wordList.count(str) == 1)
                    {
                        que.push(make_pair(str, val.second+1));
                        wordList.erase(str);
                    }
                }
            }
        }
        return 0;
    }
};