Shortest Distance from All Buildings

题目链接:https://leetcode.com/problems...

这道题要求最短的距离,一般这种要求可以到的地方的距离,都需要把整个图遍历一遍,遍历一般就是bfs和dfs。这道题不用dfs的原因是:empty的位置到building的距离是按最小值来算的,用dfs每次如果放入depth不一定是最小的距离,每次都得更新,没有效率。这道题用bfs的原因:一样的原因,因为距离就是按照最小的距离来算的,完全是bfs的思路。
visited一般两种方式:用一个boolean的矩阵,直接改写grid的值。这里用第二种。-grid[i] [j]表示(i, j)点可以reach到的building数目。当grid[i] [j] == # of buildings so far时,证明当前点还没被visited,且当前点被之前所有的buildings都visited过,那么每次bfs只访问这些点。如果该point没有被之前所有的buildings访问过,就不可能成为答案(根据要求empty的位置能到所有的buildings),其他与它相邻的点也是这样。和用boolean矩阵比,缩小了每次遍历的范围。
从每一个building,即grid[i] [j] == 1的点开始做bfs层次遍历。

  1. 用一个distance矩阵来记录(i, j)到所有building的距离和,对每一个building做bfs

  2. 每次bfs的时候,更新distance[i] [j]的值:

    • Queue<int[]> 记录point

    • 更新level += 1

    • go over当前level的全部point

      • if (i, j)在图内&grid[i] [j] = -num:

        • distance[i] [j] += level

        • grid[i] [j] --

        • q.offer(i, j)

  3. go over整个distance数组,当-grid[i] [j] == # of buildings时,更新最小的距离值

public class Solution {
    public int shortestDistance(int[][] grid) {
        /* approach: bfs, distance array
         * for each building, do a bfs, add the distance
         * variable: num: record number of buildings already searched
         * visited => use the grid => do -- if visited[i][j] = -num
         * result: the grid[i][j] == -(number of buildings) is the possible
         *         find the smallest distance[i][j]
         */
         distance = new int[grid.length][grid[0].length];
         int num = 0;
         for(int i = 0; i < grid.length; i++) {
             for(int j = 0; j < grid[0].length; j++) {
                 if(grid[i][j] == 1) {
                     bfs(grid, i, j, -num);
                     num++;
                 }
             }
         }
         int result = Integer.MAX_VALUE;
         // find the smallest distance
         for(int i = 0; i < grid.length; i++) {
             for(int j = 0; j < grid[0].length; j++) {
                 if(grid[i][j] == -num) result = Math.min(result, distance[i][j]);
             }
         }
         return result == Integer.MAX_VALUE ? -1 : result;
    }
    
    int[][] distance;
    int[] dx = new int[] {-1, 1, 0, 0};
    int[] dy = new int[] {0, 0, -1, 1};
    private void bfs(int[][] grid, int x, int y, int num) {
        Queue<int[]> q = new LinkedList();
        q.offer(new int[] {x, y});
        int len = 0;
        while(!q.isEmpty()) {
            len++;
            // current level
            for(int j = q.size(); j > 0; j--) {
                int[] cur = q.poll();
                // 4 directions
                for(int i = 0; i < 4; i++) {
                    int nx = cur[0] + dx[i], ny = cur[1] + dy[i];
                    if(nx >= 0 && nx < grid.length && ny >= 0 && ny < grid[0].length && grid[nx][ny] == num) {
                        distance[nx][ny] += len;
                        q.offer(new int[] {nx, ny});
                        grid[nx][ny]--;
                    }
                }
            }
        }
    }
}

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