Question

link

给一个矩阵，其中0代表海洋，其他数字代表高度，秉着水往低处流的原则，求出能够流向任意海洋的点。 比如说

0 0 0 1 2 3 0
0 1 2 2 4 3 2
2 1 1 3 3 2 0
0 3 3 3 2 3 3

那么就要给出 第二行的4 （这有这点出发，能够找到连通道四个0的区域的一条非递增 路线），当然也有可能找不到这样的点，或者找到多个点。

Solution

I read online and the best solution I come up with is Brute Force. I did not really understand the online discussions.

So if you are reading this and want to discuss with me, kindly leave me a comment!

Code

brute force

public void findSuperPeak(int[][] map) {
    int m = map.length;
    int n = map[0].length;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (check(map, new Pair(i, j), m, n)) {
                System.out.println("Found point (" + i + ", " + j
                        + ") with height of " + map[i][j]);
            }
        }
    }
}

private boolean check(int[][] originalMap, Pair p, int m, int n) {
    // check if point can flow to all oceans
    if (originalMap[p.x][p.y] == 0) {
        return false;
    }

    int[][] map = new int[m][n];
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            map[i][j] = originalMap[i][j];
        }
    }

    Queue<Pair> q = new LinkedList<Pair>();
    q.offer(p);

    while (!q.isEmpty()) {
        Pair top = q.poll();
        int x = top.x;
        int y = top.y;
        if (map[x][y] == -1) {
            continue;
        }
        // add neighbor nodes who are visitable from here
        if (x - 1 >= 0 && map[x - 1][y] <= map[x][y]) {
            // water can flow from:
            // 1. high altitude to lower
            // 2. from ocean to ocean
            q.offer(new Pair(x - 1, y));
        }
        if (x + 1 < m && map[x + 1][y] <= map[x][y]) {
            q.offer(new Pair(x + 1, y));
        }
        if (y - 1 >= 0 && map[x][y - 1] <= map[x][y]) {
            q.offer(new Pair(x, y - 1));
        }
        if (y + 1 < n && map[x][y + 1] <= map[x][y]) {
            q.offer(new Pair(x, y + 1));
        }

        // visit this point
        map[x][y] = -1;
    }

    // now we finished BFS and the entire map with lower altitude is visited
    // (including all ocean points). We now check if there exists a 0 in map
    for (int[] arr : map)
        for (int i : arr)
            if (i == 0) // found an unvisited ocean point
                return false;
    return true;
}