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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:
11110
11010
11000
00000
Answer: 1
Example 2:
11000
11000
00100
00011
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 }
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原文地址：http://www.cnblogs.com/reynold-lei/p/4465344.html
