LeetCode——Word Ladder-优快云博客

本文介绍了一种将一个单词通过一系列变换转化为另一个单词的方法，并确保每次变换只更改一个字符且中间产物必须存在于词典中。文章详细解释了如何将该问题转化为图上的最短路径问题，并提供了一个高效的算法实现。

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.

» Solve this problem

题目可以转换为最短路问题。

但是不能用最基本的最短路算法求解，会TLE。

需要进行一定的改变，选取了最小的节点之后，不是依次浏览更新所有节点，而是直接对该节点进行字母变换。

class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        queue<string> cur;
        set<string> set;
        cur.push(start);
        set.insert(start);
        for (int dis = 1; cur.size() > 0; dis++) {
            int count = cur.size();
            for (int i = 0; i < count; i++) {
                string q = cur.front();
                for (int j = 0; j < q.size(); j++) {
                    string temp = q;
                    for (char k = 'a'; k <= 'z'; k++) {
                        if (k != temp[j]) {
                            temp[j] = k;
                            if (temp == end) {
                                return dis + 1;
                            }
                            if (dict.find(temp) != dict.end() && set.find(temp) == set.end()) {
                                set.insert(temp);
                                cur.push(temp);
                            }
                        }
                    }
                }
                cur.pop();
            }
        }
        return 0;
    }
};