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:

    11110
    11010
    11000
    00000

    Answer: 1

    Example 2:

    11000
    11000
    00100
    00011

    Answer: 3

    Analyse: BFS or DFS. If found a '1', label that element as visited, and find all elements could be reached from that element and label them. 

    Runtime: Iteration 12ms

     1 class Solution {
     2 public:
     3     int numIslands(vector<vector<char>>& grid) {
     4         if(grid.empty() || grid[0].empty()) return 0;
     5         
     6         int result = 0;
     7         for(int i = 0; i < grid.size(); i++) {
     8             for(int j = 0; j < grid[i].size(); j++) {
     9                 if(grid[i][j] == '1') {
    10                     grid[i][j] = '2';
    11                     bfs(grid, i, j);
    12                     result++;
    13                 }
    14             }
    15         }
    16         return result;
    17     }
    18     
    19     void bfs(vector<vector<char> >& grid, int i, int j) {
    20         queue<pair<int, int> > qu;
    21         qu.push(make_pair(i, j));
    22         
    23         int m = grid.size(), n = grid[0].size();
    24         while(!qu.empty()) {
    25             pair<int, int> node = qu.front();
    26             i = node.first;
    27             j = node.second;
    28             qu.pop();
    29             // check whether its neighbour is island
    30             if(i > 0 && grid[i - 1][j] == '') {
    31                 qu.push(make_pair(i - 1, j));
    32                 grid[i - 1][j] = '2';
    33             }
    34             if(j > 0 && grid[i][j - 1] == '1') {
    35                 qu.push(make_pair(i, j - 1));
    36                 grid[i][j - 1] = '2';
    37             }
    38             if(j < n - 1 && grid[i][j + 1] == '1') {
    39                 qu.push(make_pair(i, j + 1));
    40                 grid[i][j + 1] = '2';
    41             }
    42             if(i < m - 1 && grid[i + 1][j] == '1') {
    43                 qu.push(make_pair(i + 1, j));
    44                 grid[i + 1][j] = '2';
    45             }
    46         }
    47     }
    48 };
    View Code

    Runtime: Recursive 8ms.

     1 class Solution {
     2 public:
     3     int numIslands(vector<vector<char>>& grid) {
     4         int result;
     5         if(grid.empty() || grid[0].empty()) return result;
     6         
     7         int row = grid.size(), col = grid[0].size();
     8         for(int i = 0; i < row; i++){
     9             for(int j = 0; j < col; j++){
    10                 if(grid[i][j] == '1'){
    11                     helper(grid, i, j);
    12                     result++;
    13                 }
    14             }
    15         }
    16         return result;
    17     }
    18     
    19     void helper(vector<vector<char> > &grid, int x, int y){
    20         if(x >= grid.size() || x < 0 || y >= grid[0].size() || y < 0 || grid[x][y] != '1')
    21             return;
    22             
    23         grid[x][y] = '2';
    24         helper(grid, x + 1, y);
    25         helper(grid, x - 1, y);
    26         helper(grid, x, y + 1);
    27         helper(grid, x, y - 1);
    28     }
    29 };
  • 相关阅读:
    Oracle Relink RAC Binary Options 说明
    Oracle 10g 对象 默认 ITL 数量 测试
    Oracle 相关的专业术语 说明
    Oracle 11g 新特性 自适应游标共享(Adaptive Cursor Sharing: ACS) 说明
    symbian 学习笔记(1) 基础
    symbian 学习笔记(3) 手机独有
    计算机体系结构几个常用的知识点记录
    数据结构和算法笔记(3)
    windows mobile 通用曾抽象
    一些简单常用算法整理学习
  • 原文地址:https://www.cnblogs.com/amazingzoe/p/5759327.html
Copyright © 2011-2022 走看看