Leetcode127: wordLadder I

2024年6月9日 106次阅读

Problem description

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 time.
Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
For example,

Given:

beginWord = “hit”
endWord = “cog”
wordList = [“hot”,”dot”,”dog”,”lot”,”log”,”cog”]
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.
You may assume no duplicates in the word list.
You may assume beginWord and endWord are non-empty and – are not the same.
UPDATE (2017/1/20): The wordList parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Solution

递归法（暴力搜索）：

#include <iostream>
#include <string>
#include <assert.h>
#include <vector>

using namespace std;

bool *isDealed;

bool isOneCharDiff(string s1, string s2)
{
    assert(s1.length() == s2.length());
    int cnt = 0;
    for (int i = 0; i < s1.length(); i++)
    {
        if (s1[i] != s2[i]) 
            cnt++;
    }

    if (cnt != 1)
        return false;
    return true;
}

void wordLadder(string beginWord, vector<string> &WordList, string endWord, int n, int depth, int *pL)
{
    if (beginWord == endWord)
    {
        if (*pL > depth)
            *pL = depth;
        return;
    }

    if (depth == n)
        return;

    for (int i = 0; i < n; i++)
    {
        if (!isDealed[i] && isOneCharDiff(WordList[i], beginWord))
        {
            isDealed[i] = true;
            wordLadder(WordList[i], WordList, endWord, n, depth + 1, pL);
            isDealed[i] = false;
        }
    }
}

int main()
{
    string beginWord, endWord;
    int n;
    cin >> beginWord >> endWord >> n;
    isDealed = new bool[n];
    vector<string> WordList(n);

    for (int i = 0; i < n; i++)
    {
        cin >> WordList[i];
        isDealed[i] = false;
    }

    int L = n + 1;
    wordLadder(beginWord, WordList, endWord, n, 0, &L);
    if (L < n + 1)
        cout << L << endl;
    else
        cout << 0 << endl;

    delete[] isDealed;
    return 0;
}

BFS法

#include <iostream>
#include <vector>
#include <set>
#include <string>
#include <queue>

using namespace std;

// one end bfs
/*class Solution { public: int ladderLength(string beginWord, string endWord, vector<string>& wordList) { set<string> wordSet(wordList.begin(), wordList.end()); char tmp; queue<string> Q; Q.push(beginWord); Q.push(""); //以空作为每一层的结束 int res = 1; while(!Q.empty()) { string str = Q.front(); Q.pop(); if (str != "") { for(int i = 0; i < str.length(); i++) { tmp = str[i]; for (char c = 'a'; c <= 'z'; c++) { if(str[i] == c) continue; str[i] = c; set<string>::iterator idx = wordSet.find(str); //vector<string>::iterator idx = find(wordList.begin(), wordList.end(), str); if (idx != wordSet.end()) { if (str == endWord) return res + 1; Q.push(str); wordSet.erase(idx); } } str[i] = tmp; } } else if (!Q.empty()) { res++; Q.push(""); } } return 0; } }; */

// two end bfs
class Solution {
public:
    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        set<string> wordSet(wordList.begin(), wordList.end());
        set<string> beginSet, endSet, *set1, *set2;
        beginSet.insert(beginWord);
        endSet.insert(endWord);
        int res = 1;

        if (wordSet.find(endWord) == wordSet.end())
            return 0;

        while (!beginSet.empty() && !endSet.empty())
        {
            if (beginSet.size() <= endSet.size())
            {
                set1 = &beginSet;
                set2 = &endSet;
            }
            else
            {
                set1 = &endSet;
                set2 = &beginSet;
            }
            set<string> itemSet;
            set<string>::iterator i;
            for (i = set1->begin(); i != set1->end(); i++)
            {
                string str = *i;
                for (int i = 0; i < str.length(); i++)
                {
                    char tmp = str[i];
                    for (char c = 'a'; c <= 'z'; c++)
                    {
                        if (str[i] == c) continue;
                        str[i] = c;
                        set<string>::iterator j = set2->find(str);
                        if (j != set2->end()) 
                            return res + 1;
                        set<string>::iterator idx = wordSet.find(str);
                        if (idx != wordSet.end())
                        {   
                            itemSet.insert(str);
                            wordSet.erase(idx);
                        }
                    }
                    str[i] = tmp;
                }
            }
            swap(*set1, itemSet);
            res++;
        }
        return 0;
    }
};

int main()
{
    string str_array[] ={"hot","dot","dog","lot","log","cog"}; //以这个序列为例
    string begin = "hit";
    string end = "cog";
    vector<string> wordList(str_array, str_array + 6);
    Solution s;
    cout << s.ladderLength(begin, end, wordList) << endl;
    return 0;
}

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