0079. Word Search

# 79. Word Search#

## 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.

Example:

``````board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.
``````

## 解题思路 #

• 在地图上的任意一个起点开始，向 4 个方向分别 DFS 搜索，直到所有的单词字母都找到了就输出 true，否则输出 false。

## 代码 #

``````
package leetcode

var dir = [][]int{
[]int{-1, 0},
[]int{0, 1},
[]int{1, 0},
[]int{0, -1},
}

func exist(board [][]byte, word string) bool {
visited := make([][]bool, len(board))
for i := 0; i < len(visited); i++ {
visited[i] = make([]bool, len(board[0]))
}
for i, v := range board {
for j := range v {
if searchWord(board, visited, word, 0, i, j) {
return true
}
}
}
return false
}

func isInBoard(board [][]byte, x, y int) bool {
return x >= 0 && x < len(board) && y >= 0 && y < len(board[0])
}

func searchWord(board [][]byte, visited [][]bool, word string, index, x, y int) bool {
if index == len(word)-1 {
return board[x][y] == word[index]
}
if board[x][y] == word[index] {
visited[x][y] = true
for i := 0; i < 4; i++ {
nx := x + dir[i][0]
ny := y + dir[i][1]
if isInBoard(board, nx, ny) && !visited[nx][ny] && searchWord(board, visited, word, index+1, nx, ny) {
return true
}
}
visited[x][y] = false
}
return false
}

``````

Apr 8, 2023