LeetCode HOT100 - 单词搜索

回溯

当前在找 word 的下标和当前的位置

如果不同直接退出

如果相同那么就从相邻的继续

已经被探索过的直接跳过

该位置相邻的都查看过后回溯状态

class Solution {
public:int dx[4] = {1, 0, -1, 0};int dy[4] = {0, 1, 0, -1};bool exist(vector<vector<char>>& a, string word) {int len = word.size();int n = a.size();int m = a[0].size();vector<vector<int>> vis(n, vector<int>(m, 0));auto dfs = [&](this auto&& self, int k, int x, int y) -> bool {if (a[x][y] != word[k]) {return false;}if (k == len - 1) {return true;}vis[x][y] = 1;for (int i = 0; i < 4; i++) {int xx = x + dx[i];int yy = y + dy[i];if (xx < 0 || xx >= n || yy < 0 || yy >= m) continue;if (vis[xx][yy]) continue;if (self(k + 1, xx, yy)) {vis[x][y] = 0;return true;}}vis[x][y] = 0;return false;};for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (dfs(0, i, j)) {return true;}}}return false;}
};