每日一题：二维网格图中探测环

题目

问二维矩阵中是否存在相同字母构成的环。

Solution

判环问题，显然可以想到并查集。遍历矩阵，若该点与之前遍历过的相邻点字符相同且是同一个根的话，说明存在环，否则合并这两个点。

Code

class Solution {
    static int[] pre = new int[500005];

    public int find(int x) {
        return pre[x] == x ? x : (pre[x] = find(pre[x]));
    }

    public boolean join(int x, int y) {
        int fx = find(x), fy = find(y);
        if (fx == fy) return true;
        pre[fx] = fy;
        return false;
    }

    public boolean containsCycle(char[][] grid) {
        int n = grid.length, m = grid[0].length;
        for (int i = 0; i < n * m; i++)
            pre[i] = i;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (i > 0 && grid[i][j] == grid[i - 1][j] && join(i * m + j, (i - 1) * m + j)) {
                    return true;
                }
                if (j > 0 && grid[i][j] == grid[i][j - 1] && join(i * m + j, i * m + j - 1)) {
                    return true;
                }
            }
        }
        return false;
    }
}