200. Number of Islands\127. Word Ladder

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本文介绍了两种经典的图遍历算法应用案例：岛屿数量（Number of Islands）与单词阶梯（Word Ladder）。前者通过二维网格计算相连陆地形成的岛屿数量，后者在给定词典的情况下寻找两个单词间的最短转换路径。

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Number of Islands
- 题目描述
- 代码实现
Word Ladder
- 题目描述
- 代码实现

200. Number of Islands

题目描述

Given a 2d grid map of ‘1’s (land) and ‘0’s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Answer: 1

Example 2:

Answer: 3

代码实现

法一：使用BFS计算即可：

class Solution {
public:
    int numIslands(vector<vector<char>>& grid) {
        int res = 0, m = grid.size(), n = m?grid[0].size():0;
        for(int row = 0; row < m; row++) {
            for(int col = 0; col < n; col++) {
                if(grid[row][col] == '1') {
                    res++;
                    grid[row][col] = '0';
                    queue<pair<int, int>> que;
                    que.push(make_pair(row, col));
                    while(!que.empty()) {
                        pair<int, int> tmp = que.front();
                        que.pop();
                        if(tmp.first+1 < m && grid[tmp.first+1][tmp.second] == '1') {
                            que.push(make_pair(tmp.first+1, tmp.second));
                            grid[tmp.first+1][tmp.second] = '0';
                        }  
                        if(tmp.first-1 >= 0 && grid[tmp.first-1][tmp.second] == '1') {
                            que.push(make_pair(tmp.first-1, tmp.second));
                            grid[tmp.first-1][tmp.second] = '0';
                        }
                        if(tmp.second+1 < n && grid[tmp.first][tmp.second+1] == '1') {
                            que.push(make_pair(tmp.first, tmp.second+1));
                            grid[tmp.first][tmp.second+1] = '0';
                        }
                        if(tmp.second-1 >= 0  && grid[tmp.first][tmp.second-1] == '1') {
                            que.push(make_pair(tmp.first, tmp.second-1));
                            grid[tmp.first][tmp.second-1] = '0';
                        }
                    }
                } 
            }
        }
        return res;
    }
};

127. Word Ladder

题目描述

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.

这套题目是使用BFS进行，每次把queue里面的string取出，在根据它的长度变换成别的每一种字符，把变换后的字符与目标进行匹配。若是匹配上再放回到栈中。目前这个代码实现的时候会超时。

代码实现

法一：BFS

class Solution {
public:
    bool isInWordList(vector<string> &wordList, string str) {
        for(int i = wordList.size() - 1; i >= 0; i--) 
            if(wordList[i] == str)
                return true;
        return false;        
    }

    int ladderLength(string beginWord, string endWord, vector<string>& wordList) {
        int sz = beginWord.size(), res = 0;
        queue<string> que;
        que.push(beginWord);
        string str;
        int wl = wordList.size();
        sort(wordList.begin(), wordList.end());
        while(!que.empty()) {
            if(res > wl) break;
            res++;
            int queue_size = que.size();
            for(int k = 0; k < queue_size; k++) {
                for(int i = 0; i < sz; i++) {
                    str = que.front();
                    string str2 = str;
                    for(char j = 'a'; j <= 'z'; j++) {
                        if(str[i] != j) {
                            str[i] = j;
                            if(isInWordList(wordList, str)) {
                                if(str == endWord)
                                    return res+1;
                                que.push(str);    
                            }
                        }
                    }
                }    
                que.pop();
            }    
            //cout << " in the loop  " << endl;
        }    

        return 0;
    }
};