BFS, using PriorityQueue to poll out the smallest number every time.
The largest number you get will be the result.
class Solution { public int swimInWater(int[][] grid) { if(grid==null) return 0; int m = grid.length, n = grid[0].length; boolean[][] visited = new boolean[m][n]; PriorityQueue<int[]> queue = new PriorityQueue<>((a,b)-> grid[a[0]][a[1]]-grid[b[0]][b[1]]); queue.offer(new int[]{0,0}); int[][] dirs={{-1,0},{1,0},{0,-1},{0,1}}; int res = 0; while(!queue.isEmpty()){ int[] small = queue.poll(); res = Math.max(res, grid[small[0]][small[1]]); if(small[0]==m-1 && small[1]==n-1) return res; visited[small[0]][small[1]]=true; for(int[] dir: dirs){ int x = small[0]+dir[0]; int y = small[1]+dir[1]; if(x>=0&&x<m&&y>=0&&y<n&&!visited[x][y]){ queue.offer(new int[]{x,y}); } } } return res; } }
