【路飞】算法与数据结构-被围绕的区域

不管全世界所有人怎么说，我都认为自己的感受才是正确的。无论别人怎么看，我绝不打乱自己的节奏。喜欢的事自然可以坚持，不喜欢的怎么也长久不了。

LeetCode：原题地址

题目要求

给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' ，找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。

示例 1：

输入：board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
解释：被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。

示例 2：

输入： board = [["X"]]
输出： [["X"]]

思路

并查集

• 先把四个边上的连在一起，中间的连在一起，然后根据不属于此联通的就是需要更改的位置。

/**
 * @param {character[][]} board
 * @return {void} Do not return anything, modify board in-place instead.
 */
var solve = function (board) {
    class unionFind {
        constructor(n) {
            this.count = n;
            this.parent = new Array(n);
            for (let i = 0; i < n; i++) {
                this.parent[i] = i;
            }
        }

        find(p) {
            let root = p;
            while (this.parent[root] !== root) {
                root = this.parent[root];
            }
            // 压缩路径
            while (this.parent[p] !== p) {
                let x = p;
                p = this.parent[p];
                this.parent[x] = root;
            }
            return root;
        }

        union(p, q) {
            let rootP = this.find(p);
            let rootQ = this.find(q);
            if (rootP === rootQ) return;
            this.parent[rootP] = rootQ;
            this.count--;
        }
        isConnected(x, y) {
            return this.find(x) === this.find(y)
        }
    }
    let row = board.length;
    if (row == 0) return;
    let col = board[0].length;
    let dummy = row * col;
    let uf = new unionFind(dummy);
    let arr = [[1, 0], [0, 1], [-1, 0], [0, -1]];
    for (let i = 0; i < row; i++) {
        for (let j = 0; j < col; j++) {
            if (board[i][j] == 'O') {
                if (i == 0 || j == 0 || i == row - 1 || j == col - 1) {
                    uf.union(i * col + j, dummy)
                } else {
                    //考察四个方向
                    for (let k = 0; k < 4; k++) {
                        let x = i + arr[k][0];
                        let y = j + arr[k][1];
                        if (board[x][y] == 'O') uf.union(x * col + y, i * col + j);
                    }
                }
            }
        }
    }
    for (let i = 1; i < row - 1; i++) {
        for (let j = 1; j < col - 1; j++) {
            if (!uf.isConnected(i * col + j, dummy)) board[i][j] = 'X';
        }
    }

};