两个题一样
DFS,黄金矿工是找深搜求和,更新最大值,这个题是深搜更新状态,如果满足状态继续深搜,不满足状态返回false,这条路线就不再走了,只要有一条路线符合要求,主方法中的dfs返回值就是true了。
ps:不用按单词中的顺序DFS,只要网格中的一条合法字符串包含着单词就可以。
class Solution {
int res = Integer.MIN_VALUE;
int[][] dir = new int[][] { { 0, 1 }, { 0, -1 }, { 1, 0 }, { -1, 0 } };
int r;
int c;
boolean[][] vis;
public boolean exist(char[][] board, String word) {
r = board.length;
c = board[0].length;
vis = new boolean[r][c];
for (int i = 0; i < r; i++) {
for (int j = 0; j < c; j++) {
if (board[i][j] == word.charAt(0)) {
vis[i][j] = true;
if (dfs(i, j, 1, word, board))
return true;
vis[i][j] = false;
}
}
}
return false;
}
public boolean dfs(int i, int j, int cursize, String word, char[][] board) {
if (cursize == word.length())
return true;
for (int[] d : dir) {
int newx = i + d[0];
int newy = j + d[1];
if (newx >= 0 && newx < r && newy >= 0 && newy < c && vis[newx][newy] == false) {// 在界内且未被访问
if (board[newx][newy] != word.charAt(cursize))
continue;
vis[newx][newy] = true;
if (dfs(newx, newy, cursize + 1, word, board))
return true;
vis[newx][newy] = false;
}
}
return false;
}
}