LeetCode:Word Search

标签：leetcode

题目描述:

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"]
]

思路分析：采用深度优先搜索，已经搜索过的位置用‘　’来标记。

代码：

class Solution
{
	public:
		bool search(vector<vector<char> > & board,int height,int width,int i,int j,int length,string word,int word_index)
		{
			if(board[i][j] != word[word_index] || board[i][j] == ' ')
				return false;
			if(board[i][j] == word[word_index])
			{
				if(word_index == length - 1)
					return true;
				else
				{
					char c = word[word_index];
					board[i][j] = ' ';//标记这个位置以及查找过了
					bool up = false,down = false,left = false,right = false;
					if(((i+1) < height && search(board,height,width,i+1,j,length,word,word_index+1))
						  || ((i-1) >= 0 && search(board,height,width,i-1,j,length,word,word_index+1))
						  || ((j+1) < width && search(board,height,width,i,j+1,length,word,word_index+1))
						  || ((j-1) >= 0 && search(board,height,width,i,j-1,length,word,word_index+1)))
					{
						board[i][j] = c;//回溯
						return true;
					}
					else
					{
						board[i][j]= c;
						return false;
					}
				}
			}
		}

		bool exist(vector<vector<char> > & board,string word)
		{
			int height = board.size();
			if(height == 0)
				return false;
			int width = board[0].size();
			int word_length = word.size();
			for(int i = 0;i < height;i++)
				for(int j = 0;j < width;j++)
				{
					if(search(board,height,width,i,j,word_length,word,0))
						return true;
				}
			return false;
		}
};

LeetCode:Word Search

标签：leetcode

原文地址：http://blog.csdn.net/yao_wust/article/details/45824537