Hexo

You are a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort.

A route’s effort is the maximum absolute difference in heights between two consecutive cells of the route.

Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

A*算法，启发式搜索。BFS搜索结合Priority Queue。
采用一个数组储存当前访问点的位置，以及其effort。
采用优先队列，优先搜索effort最小的方向。
每次循环倾倒出队列中所有的元素。
计算上一个节点和当前节点的差值作为nextEffort，并和上一个节点的effort作比较，较大的作为当前节点的effort，
将effort作为权重，优先搜索一个层级内effort较小的路径。
将所有操作加入队列，并排除越界的位置。
当当前节点为最后一个节点时，返回其effort。

class Solution {
    int min;
    public int minimumEffortPath(int[][] heights) {
        int[][] operations = {{1,0},{-1,0},{0,1},{0,-1}};
        int m = heights.length, n = heights[0].length;
        
        Queue<int[]> q = new PriorityQueue<>((a, b) -> a[2] - b[2]);
        int[] point = {0, 0, 0};
        int size = 1;
        int[][] visited = new int[m][n];
        
        q.add(point);
        while(!q.isEmpty()){
            
            for(int k = 0; k < size; k++){
                int[] curr = q.poll();
                int i = curr[0], j = curr[1], currEffort = curr[2];
                if(visited[i][j] == 1) continue;
                visited[i][j] = 1;
                if(i == m-1 && j == n-1) return currEffort;
                for(int[] operation : operations){
                    int nextX = i + operation[0];
                    int nextY = j + operation[1];
                    if(nextX < 0 || nextY < 0 || nextX >= m || nextY >= n) continue;
                    int nextEffort = Math.max(currEffort, Math.abs(heights[i][j] - heights[nextX][nextY]));
                    int[] next = {nextX, nextY, nextEffort};
                    
                    q.add(next);
                }
            }
            size = q.size();
        }
        return -1;
    }
}