• 695. Max Area of Island


    Given a non-empty 2D array grid of 0's and 1's, an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

    Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

    Example 1:

    [[0,0,1,0,0,0,0,1,0,0,0,0,0],
     [0,0,0,0,0,0,0,1,1,1,0,0,0],
     [0,1,1,0,1,0,0,0,0,0,0,0,0],
     [0,1,0,0,1,1,0,0,1,0,1,0,0],
     [0,1,0,0,1,1,0,0,1,1,1,0,0],
     [0,0,0,0,0,0,0,0,0,0,1,0,0],
     [0,0,0,0,0,0,0,1,1,1,0,0,0],
     [0,0,0,0,0,0,0,1,1,0,0,0,0]]
    

    Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

    Example 2:

    [[0,0,0,0,0,0,0,0]]

    Given the above grid, return 0.

    Note: The length of each dimension in the given grid does not exceed 50.

    二维矩阵里求相连的1的最大个数

    C++(22ms):

     1 class Solution {
     2 public:
     3     int AreaOfIsland(vector<vector<int>>& grid , int i , int j){
     4         if (i >= 0 && i < grid.size() && j >= 0 && j < grid[0].size() && grid[i][j] == 1){
     5             grid[i][j] = 0 ;
     6             return 1 + AreaOfIsland(grid,i+1,j)+ AreaOfIsland(grid,i-1,j)+ AreaOfIsland(grid,i,j+1)+ AreaOfIsland(grid,i,j-1) ;
     7         }
     8         return 0 ;
     9     }
    10     
    11     int maxAreaOfIsland(vector<vector<int>>& grid) {
    12         int res = 0 ;
    13         for (int i = 0 ; i < grid.size() ; i++){
    14             for (int j = 0 ; j < grid[0].size() ; j++){
    15                 if (grid[i][j] == 1){
    16                     res = max(res,AreaOfIsland(grid,i,j)) ;
    17                 }
    18             }
    19         }
    20         return res ;
    21     }
    22 };
  • 相关阅读:
    C#类型转换
    C#面向对象之多态
    C#面向对象之继承
    C#各种字段类型对比
    C#关键字:static
    C#面向对象之封装
    C#关键字:访问修饰符
    C#类型成员:构造函数
    C#类型成员:方法
    C#类类型
  • 原文地址:https://www.cnblogs.com/mengchunchen/p/7639726.html
Copyright © 2020-2023  润新知