Showing posts with label breath first search. Show all posts
Showing posts with label breath first search. Show all posts

Thursday, November 27, 2014

[LeetCode] Word Ladder I, II

Word Ladder I

Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, 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.


Word Ladder II

Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, 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"]
Return
  [
    ["hit","hot","dot","dog","cog"],
    ["hit","hot","lot","log","cog"]
  ]
Note:
  • All words have the same length.
  • All words contain only lowercase alphabetic characters.


思路：

LeetCode中为数不多的考图的难题。尽管题目看上去像字符串匹配题，但从“shortest transformation sequence from start to end”还是能透露出一点图论中最短路径题的味道。如何转化？

1. 将每个单词看成图的一个节点。
2. 当单词s1改变一个字符可以变成存在于字典的单词s2时，则s1与s2之间有连接。
3. 给定s1和s2，问题I转化成了求在图中从s1->s2的最短路径长度。而问题II转化为了求所有s1->s2的最短路径。

无论是求最短路径长度还是求所有最短路径，都是用BFS。在BFS中有三个关键步骤需要实现:

1. 如何找到与当前节点相邻的所有节点。
这里可以有两个策略：
(1) 遍历整个字典，将其中每个单词与当前单词比较，判断是否只差一个字符。复杂度为：n*w，n为字典中的单词数量，w为单词长度。
(2) 遍历当前单词的每个字符x，将其改变成a~z中除x外的任意一个，形成一个新的单词，在字典中判断是否存在。复杂度为：26*w，w为单词长度。
这里可以和面试官讨论两种策略的取舍。对于通常的英语单词来说，长度大多小于100，而字典中的单词数则往往是成千上万，所以策略2相对较优。

2. 如何标记一个节点已经被访问过，以避免重复访问。
可以将访问过的单词从字典中删除。

3. 一旦BFS找到目标单词，如何backtracking找回路径？



Word Ladder I

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class Solution {
public:
    int ladderLength(string start, string end, unordered_set<string> &dict) {
        dict.insert(end);
        queue<pair<string,int>> q;
        q.push(make_pair(start,1));
        while(!q.empty()) {
            string s = q.front().first;
            int len = q.front().second;
            if(s==end) return len;
            q.pop();
            vector<string> neighbors = findNeighbors(s, dict);
            for(int i=0; i<neighbors.size(); i++) 
                q.push(make_pair(neighbors[i],len+1));
        }
        return 0;
    }
    
    vector<string> findNeighbors(string s, unordered_set<string> &dict) {
        vector<string> ret;
        for(int i=0; i<s.size(); i++) {
            char c = s[i];
            for(int j=0; j<26; j++) {
                if(c=='a'+j) continue;
                s[i] = 'a'+j;
                if(dict.count(s)) {
                    ret.push_back(s);    
                    dict.erase(s);    
                }
            }
            s[i] = c;
        }
        return ret;
    }
};


Word Ladder II


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Monday, November 24, 2014

[LeetCode] Surrounded Regions

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.
A region is captured by flipping all 'O's into 'X's in that surrounded region.
For example,
X X X X
X O O X
X X O X
X O X X
After running your function, the board should be:
X X X X
X X X X
X X X X
X O X X



思路：

这题题意也略显含糊，实际上就是要将所有以O组成、但不连通到网格边缘的区域变为X。所以我们可以先在四边上寻找连通到边缘的区域，将它们的O都变成Y。剩余的所有O一定无法连通到边缘，所以可以全部变为X。最后再将所有Y变回O。

这题的测试比较tricky，上来没多想就用递归的DFS实现flood fill，结果发现run time error，栈溢出了：


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class Solution {
public:
    void solve(vector<vector<char>> &board) {
        if(board.size()<3 || board[0].size()<3) return;
        fillBorders(board, 'O', 'Y');
        replace(board, 'O', 'X');
        fillBorders(board, 'Y', 'O');
    }
    
    void fill(vector<vector<char>> &board, int i, int j, char target, char c) {
        int m = board.size(), n = board[0].size();
        if(i<0 || j<0 || i>=m || j>=n || board[i][j]!=target) return;
        board[i][j] = c;
        fill(board, i-1, j, target, c);
        fill(board, i+1, j, target, c);
        fill(board, i, j-1, target, c);
        fill(board, i, j+1, target, c);
    }
    
    void fillBorders(vector<vector<char>> &board, char target, char c) {
        int m = board.size(), n = board[0].size();
        for(int i=0; i<m; i++) {
            if(board[i][0]==target) fill(board, i, 0, target, c);
            if(board[i][n-1]==target) fill(board, i, n-1, target, c);
        }
        
        for(int j=1; j<n-1; j++) {
            if(board[0][j]==target) fill(board, 0, j, target, c);
            if(board[m-1][j]==target) fill(board, m-1, j, target, c);
        }
    }
    
    
    void replace(vector<vector<char>> &board, char target, char c) {
        int m = board.size(), n = board[0].size();
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(board[i][j]==target)
                    board[i][j] = c;
            }
        }
    }
};


于是些了个迭代版的DFS，果然过了大数据测试。当然这里也可以用BFS实现flood fill。

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class Solution {
public:
    void solve(vector<vector<char>> &board) {
        if(board.size()<3 || board[0].size()<3) return;
        fillBorders(board, 'O', 'Y');
        replace(board, 'O', 'X');
        fillBorders(board, 'Y', 'O');
    }
    
    void fill(vector<vector<char>> &board, int i, int j, char target, char c) {
        int m = board.size(), n = board[0].size();
        if(i<0 || j<0 || i>=m || j>=n || board[i][j]!=target) return;
        stack<pair<int,int>> s;
        s.push(make_pair(i,j));
        
        while(!s.empty()) {
            i = s.top().first;
            j = s.top().second;
            s.pop();
            board[i][j] = c;
            if(i>0 && board[i-1][j]==target) s.push(make_pair(i-1,j));
            if(i<m-1 && board[i+1][j]==target) s.push(make_pair(i+1,j));
            if(j>0 && board[i][j-1]==target) s.push(make_pair(i,j-1));
            if(j<n-1 && board[i][j+1]==target) s.push(make_pair(i,j+1));
        }
    }
    
    void fillBorders(vector<vector<char>> &board, char target, char c) {
        int m = board.size(), n = board[0].size();
        for(int i=0; i<m; i++) {
            if(board[i][0]==target) fill(board, i, 0, target, c);
            if(board[i][n-1]==target) fill(board, i, n-1, target, c);
        }
        
        for(int j=1; j<n-1; j++) {
            if(board[0][j]==target) fill(board, 0, j, target, c);
            if(board[m-1][j]==target) fill(board, m-1, j, target, c);
        }
    }
    
    
    void replace(vector<vector<char>> &board, char target, char c) {
        int m = board.size(), n = board[0].size();
        for(int i=0; i<m; i++) {
            for(int j=0; j<n; j++) {
                if(board[i][j]==target)
                    board[i][j] = c;
            }
        }
    }
};



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Thursday, November 20, 2014

[LeetCode] Clone Graph

Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.

OJ's undirected graph serialization:
Nodes are labeled uniquely.
We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.
As an example, consider the serialized graph {0,1,2#1,2#2,2}.
The graph has a total of three nodes, and therefore contains three parts as separated by #.
  1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
  2. Second node is labeled as 1. Connect node 1 to node 2.
  3. Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.
Visually, the graph looks like the following:
       1
      / \
     /   \
    0 --- 2
         / \
         \_/


思路

和Copy List with Random Pointer那题的思路一样。用一个hash table记录原图节点和复制图节点间的对应关系，以防止重复建立节点。和那题的不同在于遍历原图相对比linked list的情况复杂一点。可以用BFS或DFS来遍历原图。而hash table本身除了记录对应关系外，还有记录原图中每个节点是否已经被visit的功能。

BFS遍历：

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class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(!node) return NULL;
        UndirectedGraphNode *p1 = node;
        UndirectedGraphNode *p2 = new UndirectedGraphNode(node->label);
        unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> ht;
        queue<UndirectedGraphNode*> q;        
        q.push(node);
        ht[node] = p2;
        
        while(!q.empty()) {
            p1 = q.front();
            p2 = ht[p1];
            q.pop();
            for(int i=0; i<p1->neighbors.size(); i++) {
                UndirectedGraphNode *nb = p1->neighbors[i];
                if(ht.count(nb)) {
                    p2->neighbors.push_back(ht[nb]);
                }
                else {
                    UndirectedGraphNode *temp = new UndirectedGraphNode(nb->label);
                    p2->neighbors.push_back(temp);
                    ht[nb] = temp;
                    q.push(nb);
                }
            }
        }
        
        return ht[node];
    }
};


DFS遍历：

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class Solution {
public:
    UndirectedGraphNode *cloneGraph(UndirectedGraphNode *node) {
        if(!node) return NULL;
        unordered_map<UndirectedGraphNode*, UndirectedGraphNode*> ht;
        stack<UndirectedGraphNode*> s;        
        s.push(node);
        ht[node] = new UndirectedGraphNode(node->label);
        
        while(!s.empty()) {
            UndirectedGraphNode *p1 = s.top(), *p2 = ht[p1];
            s.pop();
            
            for(int i=0; i<p1->neighbors.size(); i++) {
                UndirectedGraphNode *nb = p1->neighbors[i];
                if(ht.count(nb)) {
                    p2->neighbors.push_back(ht[nb]);
                }
                else {
                    UndirectedGraphNode *temp = new UndirectedGraphNode(nb->label);
                    p2->neighbors.push_back(temp);
                    ht[nb] = temp;
                    s.push(nb);
                }
            }
        }
        
        return ht[node];
    }
};
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