leetcode417

DFS思路

class Solution {
    private int m, n;
    private int[][] matrix;
    private int[][] direction = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

    public List<int[]> pacificAtlantic(int[][] matrix) {
        List<int[]> ret = new ArrayList<>();
        if (matrix == null || matrix.length == 0) {
            return ret;
        }

        m = matrix.length;
        n = matrix[0].length;
        this.matrix = matrix;
        boolean[][] canReachP = new boolean[m][n];
        boolean[][] canReachA = new boolean[m][n];

        for (int i = 0; i < m; i++) {
            dfs(i, 0, canReachP);//从左边界（太平洋）开始，即第0列
            dfs(i, n - 1, canReachA);//从右边界（大西洋）开始，即最后一列
        }
        for (int i = 0; i < n; i++) {
            dfs(0, i, canReachP);//从上边界（太平洋）开始，即第0行
            dfs(m - 1, i, canReachA);//从下边界（大西洋）开始，即最后一行
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (canReachP[i][j] && canReachA[i][j]) {//同时能从太平洋边界和大西洋边界达到的点（求交集）
                    ret.add(new int[]{i, j});
                }
            }
        }

        return ret;
    }

    private void dfs(int r, int c, boolean[][] canReach) {
        if (canReach[r][c]) {
            return;
        }
        canReach[r][c] = true;
        for (int[] d : direction) {//遍历上下左右，四个方向
            int nextR = d[0] + r;
            int nextC = d[1] + c;
            if (nextR < 0 || nextR >= m || nextC < 0 || nextC >= n//越界（非法）
                    || matrix[r][c] > matrix[nextR][nextC]) {//从高处（边界）流向低处（中央区域）（非法）

                continue;
            }
            dfs(nextR, nextC, canReach);
        }
    }
}