给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
输出：true
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
输出：true
输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
输出：false
提示：
board 和 word 仅由大小写英文字母组成
递归的关键点
那么，哪些情况说明这是一个错的点：
class Solution {
public boolean exist(char[][] board, String word) {
if (board == null || board.length == 0) {
return false;
}
boolean[][] visited = new boolean[board.length][board[0].length];
char[] chars = word.toCharArray();
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (existHelper(board, visited, chars, i, j, 0)) {
return true;
}
}
}
return false;
}
private boolean existHelper(char[][] board, boolean[][] visited, char[] chars, int row, int column, int index) {
if (index == chars.length) {
return true;
}
int[][] direction = new int[][]{
{0, 1},
{1, 0},
{0, -1},
{-1, 0}
};
if (row >= 0 && row < board.length &&
column >= 0 && column < board[0].length &&
board[row][column] == chars[index] &&
!visited[row][column]) {
visited[row][column] = true;
for (int[] dir : direction) {
int newX = row + dir[0];
int newY = column + dir[1];
if (existHelper(board, visited, chars, newX, newY, index + 1)) {
return true;
}
}
visited[row][column] = false;
}
return false;
}
}
时间复杂度：O（M*N * 3^L）
空间复杂度：O（M*N）
思路同上
func exist(board [][]byte, word string) bool {
m, n := len(board), len(board[0])
used := make([][]bool, m)
for i := 0; i < m; i++ {
used[i] = make([]bool, n)
}
var canFind func(r, c, i int) bool
canFind = func(r, c, i int) bool {
if i == len(word) {
return true
}
if r < 0 || r >= m || c < 0 || c >= n {
return false
}
if used[r][c] || board[r][c] != word[i] {
return false
}
used[r][c] = true
canFindRest := canFind(r+1, c, i+1) || canFind(r-1, c, i+1) ||
canFind(r, c+1, i+1) || canFind(r, c-1, i+1)
if canFindRest {
return true
} else {
used[r][c] = false
return false
}
}
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if board[i][j] == word[0] && canFind(i, j, 0) {
return true
}
}
}
return false
}
时间复杂度：O（M*N * 3^L）
空间复杂度：O（M*N）
