LeetCode刷题【Array】 Word Search

最新推荐文章于 2022-12-30 11:28:20 发布

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本文介绍了一种算法，用于判断一个给定的单词是否能在二维字符网格中找到路径构成该单词。通过递归回溯的方式检查所有可能的路径，并考虑了上下左右四个方向的移动。

题目：

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 .

解决方法一：Runtime: 14 ms

public class Solution {
	    public boolean exist(char[][] board, String word) {	
			if(null==board) return false;
			if(null==word) return true;
			int [][] roadMap = new int[board.length][];
			for(int i=0;i<board.length;i++){
				roadMap[i] = new int[board[i].length];
			}
			for(int i=0;i<board.length;i++){
				for(int j=0;j<board[i].length;j++){
					if(board[i][j]==word.charAt(0)){
						roadMap[i][j] = 1;
						boolean flag = backTraceWord(board,i,j,roadMap,word,1);
						if(flag==true) return true;
						roadMap[i][j] = 0;
					}
				}
			}
	        return false;
	    }
		
		private boolean backTraceWord(char[][] board, int r,int c, int[][] roadMap, String word, int index){
			if(index>=word.length()) return true;
			boolean flag = false;
			if(c-1>=0&&roadMap[r][c-1]!=1&&board[r][c-1]==word.charAt(index)){
				roadMap[r][c-1]=1;
				flag|=backTraceWord(board,r,c-1,roadMap,word,index+1);
				if(flag==true) return true;
				roadMap[r][c-1]=0;
			}
			if(r-1>=0&&roadMap[r-1][c]!=1&&board[r-1][c]==word.charAt(index)){
				roadMap[r-1][c]=1;
				flag|=backTraceWord(board,r-1,c,roadMap,word,index+1);
				if(flag==true) return true;
				roadMap[r-1][c]=0;
			}
			if(c+1<roadMap[r].length&&roadMap[r][c+1]!=1&&board[r][c+1]==word.charAt(index)){
				roadMap[r][c+1]=1;
				flag|=backTraceWord(board,r,c+1,roadMap,word,index+1);
				if(flag==true) return true;
				roadMap[r][c+1]=0;
			}
			if(r+1<roadMap.length&&roadMap[r+1][c]!=1&&board[r+1][c]==word.charAt(index)){
				roadMap[r+1][c]=1;
				flag|=backTraceWord(board,r+1,c,roadMap,word,index+1);
				if(flag==true) return true;
				roadMap[r+1][c]=0;
			}
			return flag;			
		}
	}

解决方法二：不用额外空间

public boolean exist(char[][] board, String word) {
    char[] w = word.toCharArray();
    for (int y=0; y<board.length; y++) {
    	for (int x=0; x<board[y].length; x++) {
    		if (exist(board, y, x, w, 0)) return true;
    	}
    }
    return false;
}

private boolean exist(char[][] board, int y, int x, char[] word, int i) {
	if (i == word.length) return true;
	if (y<0 || x<0 || y == board.length || x == board[y].length) return false;
	if (board[y][x] != word[i]) return false;
	board[y][x] ^= 256;
	boolean exist = exist(board, y, x+1, word, i+1)
		|| exist(board, y, x-1, word, i+1)
		|| exist(board, y+1, x, word, i+1)
		|| exist(board, y-1, x, word, i+1);
	board[y][x] ^= 256;
	return exist;
}

参考：

【1】https://leetcode.com/