Number of Islands

最新推荐文章于 2025-05-07 17:27:01 发布

转载最新推荐文章于 2025-05-07 17:27:01 发布 · 80 阅读

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原文链接：http://www.cnblogs.com/reynold-lei/p/4465344.html

本文讨论了如何优化解决二维网格地图中岛屿数量计算的问题。通过改变地图状态并在队列中添加岛屿前先进行修改，避免了重复计算，提高了算法效率。详细介绍了改进后的算法步骤，并附带了实例分析。

Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Answer: 1

Example 2:

Answer: 3

 1 public class Solution {
 2     class Node{
 3         int x;
 4         int y;
 5         public Node(int x, int y){
 6             this.x = x;
 7             this.y = y;
 8         }
 9     }
10     public int numIslands(char[][] grid) {
11         if(grid == null || grid.length == 0 || grid[0].length == 0) return 0;
12         int result = 0;
13         for(int i = 0; i < grid.length; i ++){
14             for(int j = 0; j < grid[0].length; j ++){
15                 if(grid[i][j] == '1'){
16                     result ++;
17                     //set this island to be water ('1' to '0')
18                     LinkedList<Node> queue = new LinkedList<Node>();
19                     queue.add(new Node(i, j));
20                     while(!queue.isEmpty()){
21                         Node tmp = queue.poll();
22                         grid[tmp.x][tmp.y] = '0';
23                         if(tmp.x > 0 && grid[tmp.x - 1][tmp.y] == '1') queue.add(new Node(tmp.x - 1, tmp.y));
24                         if(tmp.x < grid.length - 1 && grid[tmp.x + 1][tmp.y] == '1') queue.add(new Node(tmp.x + 1, tmp.y));
25                         if(tmp.y > 0 && grid[tmp.x][tmp.y - 1] == '1') queue.add(new Node(tmp.x, tmp.y - 1));
26                         if(tmp.y < grid[0].length - 1 && grid[tmp.x][tmp.y + 1] == '1') queue.add(new Node(tmp.x, tmp.y + 1));
27                     }
28                 }
29             }
30         }
31         return result;
32     }
33 }

Submission Result: Time Limit ExceededMore Details

Last executed input:
"11111011111111101011",
"01111111111110111110",
"10111001101111111111",
"11110111111111111111",
"10011111111111111111",
"10111111011101110111",
"01111111111101101111",
"11111111111101111011",
"11111111110111111111",
"11111111111111111111",
"01111111011111111111",
"11111111111111111111",
"11111111111111111111",
"11111011111110111111",
"10111110111011110111",
"11111111111101111110",
"11111111111110111100",
"11111111111111111111",
"11111111111111111111",
"11111111111111111111".

原因是很多点被重复放到了queue中。Because when a Point is added, the state of this Point should be changed. In your code, one Point may be add in the LinkedList more than once.

So next solution:

modify grid[i][j] firstly (before put into the queue).

 1 public class Solution {
 2     class Node{
 3         int x;
 4         int y;
 5         public Node(int x, int y){
 6             this.x = x;
 7             this.y = y;
 8         }
 9     }
10     public int numIslands(char[][] grid) {
11         if(grid == null || grid.length == 0 || grid[0].length == 0) return 0;
12         int result = 0;
13         for(int i = 0; i < grid.length; i ++){
14             for(int j = 0; j < grid[0].length; j ++){
15                 if(grid[i][j] == '1'){
16                     result ++;
17                     //set this island to be water ('1' to '0')
18                     LinkedList<Node> queue = new LinkedList<Node>();
19                     queue.add(new Node(i, j));
20                     grid[i][j] = '0';
21                     while(!queue.isEmpty()){
22                         Node tmp = queue.poll();
23                         if(tmp.x > 0 && grid[tmp.x - 1][tmp.y] == '1'){
24                             grid[tmp.x - 1][tmp.y] = '0';
25                             queue.add(new Node(tmp.x - 1, tmp.y));  
26                         } 
27                         if(tmp.x < grid.length - 1 && grid[tmp.x + 1][tmp.y] == '1'){
28                             grid[tmp.x + 1][tmp.y] = '0';
29                             queue.add(new Node(tmp.x + 1, tmp.y));
30                         } 
31                         if(tmp.y > 0 && grid[tmp.x][tmp.y - 1] == '1'){
32                             grid[tmp.x][tmp.y - 1] = '0';
33                             queue.add(new Node(tmp.x, tmp.y - 1));
34                         } 
35                         if(tmp.y < grid[0].length - 1 && grid[tmp.x][tmp.y + 1] == '1'){
36                             grid[tmp.x][tmp.y + 1] = '0';
37                             queue.add(new Node(tmp.x, tmp.y + 1));
38                         } 
39                     }
40                 }
41             }
42         }
43         return result;
44     }
45 }

转载于:https://www.cnblogs.com/reynold-lei/p/4465344.html