LeetCode 题解（94）: Word Ladder

题目：

Given two words (beginWord and endWord), and a dictionary, find the length of shortest transformation sequence from beginWord to endWord, such that:

1. Only one letter can be changed at a time
2. 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.

题解：

用BFS + 单Queue + 每次与wordDict比较，C++可以刚刚通过时间测试，java不能通过时间测试。于是网上学来BFS + 双Queue + 每次修改一个字符并与wordDict比较，每次止血比较 stringLength * 26次。

C++版：

class Solution {
public:
    int ladderLength(string beginWord, string endWord, unordered_set<string>& wordDict) {
        if(beginWord == endWord)
            return 2;
        if(wordDict.empty())
            return 0;
        wordDict.insert(endWord);
        queue<pair<string, int>> forward;
        forward.push(pair<string, int>(beginWord, 1));
        if(wordDict.find(beginWord) != wordDict.end())
            wordDict.erase(beginWord);
        while(!forward.empty()) {
            pair<string, int> front = forward.front();
            forward.pop();
            if(front.first == endWord)
                return front.second;
            for(unordered_set<string>::iterator iter = wordDict.begin(); iter != wordDict.end(); ) {
                if(distance(*iter, front.first)) {
                    forward.push(pair<string, int>(*iter, front.second+1));
                    wordDict.erase(iter++);
                } else {
                    iter++;
                }
            }
            cout << endl;
        }
        return 0;
    }
    
    bool distance(const string& a, const string& b) {
        int diff = 0;
        for(unsigned int i = 0; i < a.length(); i++) {
            if(a[i] != b[i])
                diff++;
            if(diff > 1)
                return false;
        }
        if(diff == 1)
            return true;
        else
            return false;
    }
};

Java版：

注意String比较的时候要用.equals()

public class Solution {
    public int ladderLength(String beginWord, String endWord, Set<String> wordDict) {
        if(beginWord == null || endWord == null || wordDict.size() == 0)
            return 0;

        Queue<String> queue1 = new LinkedList<>();
        Queue<String> queue2 = new LinkedList<>();
        queue1.add(beginWord);
        int distance = 1;
        while(wordDict.size() != 0 && queue1.size() != 0) {
            while(queue1.size() != 0) {
                String current = queue1.poll();
                for(int i = 0; i < current.length(); i++) {
                    for(char j = 'a'; j <= 'z'; j++) {
                        String temp = current;
                        if(current.charAt(i) == j)
                            continue;
                        StringBuilder newString = new StringBuilder(current);
                        newString.setCharAt(i, j);
                        current = newString.toString();
                        if(current.equals(endWord))
                            return distance + 1;
                        if(wordDict.contains(current)) {
                            queue2.add(current);
                            wordDict.remove(current);
                        }
                        current = temp;
                    }
                }
            }
            distance += 1;
            queue1.addAll(queue2);
            queue2.clear();
        }
        return 0;
    }
}

Python版：

注意复制deque的时候要用深copy。import copy

a = copy.copy(b)

from collections import deque

class Solution:
    # @param beginWord, a string
    # @param endWord, a string
    # @param wordDict, a set<string>
    # @return an integer
    def ladderLength(self, beginWord, endWord, wordDict):
        if beginWord == None or endWord == None or len(wordDict) == 0:
            return 0
            
        q1 = deque()
        q1.append(beginWord)
        q2 = deque()
        distance = 1
        while len(wordDict) != 0 and len(q1) != 0:
            while len(q1) != 0:
                current = q1.pop()
                for i in range(len(current)):
                    temp = current
                    for j in "abcdefghijklmnopqrstuvwxyz":
                        if j == i:
                            continue
                        current = current[0:i] + j + current[i+1:]
                        if current == endWord:
                            return distance + 1
                        if current in wordDict:
                            q2.append(current)
                            wordDict.remove(current)
                        current = temp
            distance += 1
            q1 = copy.copy(q2)
            q2.clear()
            
        return 0