zoukankan html css js c++ java

Number of Islands

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 }

查看全文

相关阅读:
LOJ-10096（强连通+bfs）
LOJ-10095（缩点的特殊使用）
LOJ-10094（强连通分量）
LOJ-10092（最大半连通子图）
【BZOJ3489】A simple rmq problem（KD-Tree）
UVA10384 推门游戏 The Wall Pushers（IDA*）
[SCOI2005]骑士精神（IDA*）
浅谈A*算法
 【模板】K-D Tree
【XSY1953】【BZOJ4012】【HNOI2015】开店（动态点分治）

原文地址：https://www.cnblogs.com/reynold-lei/p/4465344.html