LeetCode-Word Search

最新推荐文章于 2019-04-18 11:37:30 发布

原创最新推荐文章于 2019-04-18 11:37:30 发布 · 176 阅读

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本文介绍了一种在二维网格中查找指定单词的算法实现。通过递归方式从每个格子出发，检查单词是否能由相邻格子的字符构成。文章包含完整的C++代码示例，演示了如何使用方向数组来遍历相邻单元格，并通过布尔矩阵记录已访问过的单元格。

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 =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,

word = "ABCB", -> returns false.

一道搜索题，只不过这道题要以每个点为起点开始搜，

AC：

#include <iostream>
#include <string>
#include <vector>
using namespace std;

 

int nextt[4][2] = {{-1,0},{0,1},{1,0},{0,-1}};//上右下左
int n,m;
vector<vector<bool>> visited;
//判断边界
bool inAway( int x , int y ){
	return ( x>=0 && x < m && y >= 0 && y < n );
}
bool searchWord( vector<vector<char>>& board, string word , int index , int startx , int starty ){

	if( index == word.size() - 1 )
		return word[index] == board[startx][starty];

        //因为单词是连续的  只有当前位置的字母和字母位置的索引相同  才能寻找下一个
	if( board[startx][starty] == word[index] ){

		visited[startx][starty] = true;
                
		for( int i = 0 ; i < 4 ; i ++ ){
			int newx = startx + nextt[i][0];
			int newy = starty + nextt[i][1];
			if( inAway( newx,newy ) && !visited[newx][newy] )
				if( searchWord( board , word , index+1 , newx , newy ) ) 
					return true;
		} 

		visited[startx][starty] = false;
	}
	return false;
}


bool exist(vector<vector<char>>& board, string word) {

	m = board.size();
	n = board[0].size();
	visited = vector<vector<bool> >(m,vector<bool>(n,false));
	for( int i = 0 ; i < board.size() ; i ++ )
		for( int j = 0 ; j < board[i].size() ; j ++ ){
			//从地图的每一个点当做起点开始搜  
			if( searchWord( board , word , 0 , i , j ) )
				return true;

		}

		return false;

}

int main()
{
	string s = "ABCB";
	vector<vector<char>> board;
	vector<char> v;
	
	v.push_back('A');
	v.push_back('B');
	v.push_back('C');
	v.push_back('E');
	board.push_back(v);
	v.clear();

	v.push_back('S');
	v.push_back('F');
	v.push_back('C');
	v.push_back('S');
	board.push_back(v);
	v.clear();

	v.push_back('A');
	v.push_back('D');
	v.push_back('E');
	v.push_back('E');
	board.push_back(v);
	v.clear();


	cout << exist(board,s) << endl; 
	return 0;
}