标签：leetcode   算法   

问题描述：

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

思路：

回溯法，找到一个可行解即可。

此题根据每个字符位置，找和它邻近的位置(上下左右)是否是word的下一个字符。每次考虑这四种情况。

同时本题中要求每个相同位置的字符只能用一次。所以要用记事本record记录已经用了哪些位置的字符。

代码：

class Solution { //C++
public:
    bool exist(vector<vector<char> > &board, string word) {
        if(board.size() == 0)
            return false;
        if(word == "")
            return true;
            
        vector<vector<int> > record;
        for(int i = 0; i < board.size(); i++){
            vector<int > tmp ;
            for(int j = 0; j < board[0].size(); j++)
                tmp.push_back(0);
            record.push_back(tmp);
                
        }
        
        
        int size = word.size();
        for(int i = 0; i < board.size(); i++){
            for(int j = 0; j < board[0].size(); j++){
                if(board[i][j] == word[0]){
                    record[i][j] = 1;
                    if(isExist(board, word.substr(1,size-1),i,j,record))
                        return true;
                    record[i][j] = 0;
                }
            }
        }
        return false;
    }
    bool isExist(vector<vector<char> > &board, string word, int row, int col,vector<vector<int> > &record){
        if(word.size() == 0){
            return true;
        }
        // up down left right
        bool up = false , down = false, left = false, right =false;
        int size = word.size();
        int frow = 0, fcol = 0, erow = board.size()-1, ecol = board[0].size()-1;
        
        // up
        if(row != frow && record[row-1][col] == 0 && word[0] == board[row-1][col]){
            if(size == 1)
                return true;
            else {
                record[row-1][col] = 1;
                up = isExist(board,word.substr(1,size-1),row-1,col,record);
                record[row-1][col] = 0;
            }
            if(up == true)
                return true;
        }
        
        // down
        if(row != erow && record[row+1][col] == 0 &&word[0] == board[row+1][col]){
            if(size == 1)
                return true;
            else {
                record[row+1][col] = 1;
                down = isExist(board,word.substr(1,size-1),row+1,col,record);
                record[row+1][col] = 0;
            }
            if(down == true)
                return true;
        }
        
        // left
        if(col != fcol && record[row][col-1] == 0 &&word[0] == board[row][col-1]){
            if(size == 1)
                return true;
            else {
                record[row][col-1] = 1;
                left = isExist(board,word.substr(1,size-1),row,col-1,record);
                record[row][col-1] = 0;
            }
            if(left == true)
                return true;
        }
        
        // right
        if(col != ecol && record[row][col+1] == 0 &&word[0] == board[row][col+1]){
            if(size == 1)
                return true;
            else {
                record[row][col+1] = 1;
                right = isExist(board,word.substr(1,size-1),row,col+1,record);
                record[row][col+1] = 0;
            }
            if(right == true)
                return true;
        }
        
        return false;
    }
};

[leetcode]Word Search

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/40833275