LeetCode127 Word Ladder

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详细见：leetcode.com/problems/word-ladder

Java Solution: github

package leetcode;import java.util.ArrayList;import java.util.HashSet;import java.util.LinkedList;import java.util.Queue;/* * Given two words (beginWord and endWord), and a dictionary's word list, *  find the length of shortest transformation sequence *   from beginWord to endWord, such that:Only one letter can be changed at a timeEach intermediate word must exist in the word listFor 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. */import java.util.Set;public class P127_WordLadder {/* * 184 ms * 6.20% */static class Solution_Answer {long count = 0;long count_replace = 0;    public int ladderLength(String start, String end, Set<String> dict) {        if (dict == null) {            return 0;        }        if (start.equals(end)) {            return 1;        }        dict.add(start);        dict.add(end);        HashSet<String> hash = new HashSet<String>();        Queue<String> queue = new LinkedList<String>();        queue.offer(start);        hash.add(start);        int length = 1;        while(!queue.isEmpty()) {            length++;            int size = queue.size();            for (int i = 0; i < size; i++) {                String word = queue.poll();                for (String nextWord: getNextWords(word, dict)) {                    if (hash.contains(nextWord)) {                        continue;                    }                    if (nextWord.equals(end)) {                    System.out.println("count : " + count);                    System.out.println("count_replace : " + count_replace);                        return length;                    }                    count ++;                    hash.add(nextWord);                    queue.offer(nextWord);                }            }        }        return 0;    }    // replace character of a string at given index to a given character    // return a new string    private String replace(String s, int index, char c) {        char[] chars = s.toCharArray();        chars[index] = c;        count_replace ++;        return new String(chars);    }        // get connections with given word.    // for example, given word = 'hot', dict = {'hot', 'hit', 'hog'}    // it will return ['hit', 'hog']    private ArrayList<String> getNextWords(String word, Set<String> dict) {        ArrayList<String> nextWords = new ArrayList<String>();        for (char c = 'a'; c <= 'z'; c++) {            for (int i = 0; i < word.length(); i++) {                if (c == word.charAt(i)) {                    continue;                }                String nextWord = replace(word, i, c);                if (dict.contains(nextWord)) {                    nextWords.add(nextWord);                }            }        }        return nextWords;    }}}

C Solution: github

#pragma warning(disable:4786)#pragma warning(disable:4503)/*    url: leetcode.com/problems/word-ladder    need to use advanced data structure    choose cpp    AC 323ms 29.09%*/#include <iostream>#include <string>#include <vector>#include <map>#include <queue>#include <set>using namespace std;int ladderLength(string s, string t, vector<string >& w) {    int sn = s.size() , i;    set<string > nv , hv;    for (i = 0; i < w.size(); i ++) nv.insert(w[i]);    map<string, vector<string > > m;    queue<string > q;    nv.insert(s);    q.push(s);    bool isFind = false;    int ans = 2;    while (! q.empty()) {        int size = q.size();        while (size -- > 0) {            string n = q.front();            string v = n;            q.pop();            for (i = 0; i < sn; i ++) {                for (char c = 'a'; c <= 'z'; c ++) {                    v[i] = c;                    if (! nv.count(v)) continue;                    if (! hv.count(v)) {                        hv.insert(v);                        q.push(v);                    }                    m[v].push_back(n);                    if (! v.compare(t)) isFind = true;                }                v[i] = n[i];            }        }        if (isFind) break;        ans ++;        for (set<string >::iterator iter = hv.begin(); iter != hv.end(); iter ++)            nv.erase(*iter);        hv.clear();    }    return isFind ? ans : 0;;}int main() {    string g[] = {"hot","dot","dog","lot","log","cog"};    string s = "hit", t = "cog";    vector<string > w;    int i;    for (i = 0; i < (sizeof(g)/sizeof(g[0])); i ++) w.push_back(g[i]);    int ans = ladderLength(s, t, w);    cout<<ans<<endl;    return 0;}

Python Solution: github

#coding=utf-8'''    url: leetcode.com/problems/word-ladder    @author:     zxwtry    @email:      zxwtry@qq.com    @date:       2017年5月12日    @details:    Solution:  902ms 44.52%'''from jinja2._compat import unichrclass Solution(object):    def ladderLength(self, b, e, w):        """        :type b: str        :type e: str        :type w: List[str]        :rtype: int        """        nv = set(w)        hv = set()        sn = len(b)        q = []        nv.add(b)        q.append(b)        isFind = False        count = 2        while True:            l = len(q)            if l == 0: break            while l > 0:                l -= 1                s = q[0]                q.remove(q[0])                cs = [s[i] for i in range(sn)]                for i in range(sn):                    for j in range(26):                        cs[i] = unichr(97+j)                        ns = "".join(cs)                        if not ns in nv: continue                        if not ns in hv:                            hv.add(ns)                            q.append(ns)                        if ns == e: isFind = True                    cs[i] = s[i]            if isFind: break            count += 1            for v in hv: nv.remove(v)            hv.clear()        return 0 if not isFind else count        if __name__ == "__main__":    b, e = "hit", "cog"    w = ["hot","dot","hog","dog","lot","log","cog"]#     b, e = "a", "c"#     w = ["a", "b", "c"]    print(Solution().ladderLength(b, e, w))

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