Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. Return the answer in any order.
A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.
Given an m x n grid of characters board and a string word, return trueifwordexists in the grid.
The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.
Example 1:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true
Example 2:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true
Example 3:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false
Constraints:
m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board and word consists of only lowercase and uppercase English letters.
Follow up: Could you use search pruning to make your solution faster with a larger board?
Solution
Approach #0
var ( dx = []int{0, 1, 0, -1} dy = []int{1, 0, -1, 0})funcexist(board [][]byte, word string) bool { m, n :=len(board), len(board[0]) vis :=make([][]bool, m)for i :=range vis { vis[i] =make([]bool, n) }var b func(x, y, index int) bool b =func(x, y, index int) bool {if board[x][y] != word[index] {returnfalse }if index ==len(word)-1 {returntrue } vis[x][y] =truedeferfunc() { vis[x][y] =false }()for i :=0; i <4; i++ { xx, yy := x+dx[i], y+dy[i]if xx >=0&& xx < m && yy >=0&& yy < n &&!vis[xx][yy] {if b(xx, yy, index+1) {returntrue } } }returnfalse }for i, row :=range board {for j :=range row {if b(i, j, 0) {returntrue } } }returnfalse}