[leetcode]827. Making A Large Island

链接:https://leetcode.com/problems/making-a-large-island/description/

In a 2D grid of 0s and 1s, we change at most one 0 to a 1.

After, what is the size of the largest island? (An island is a 4-directionally connected group of 1s).

Example 1:

Input: [[1, 0], [0, 1]]
Output: 3
Explanation: Change one 0 to 1 and connect two 1s, then we get an island with area = 3.

Example 2:

Input: [[1, 1], [1, 0]]
Output: 4
Explanation: Change the 0 to 1 and make the island bigger, only one island with area = 1.

Example 3:

Input: [[1, 1], [1, 1]]
Output: 4
Explanation: Can't change any 0 to 1, only one island with area = 1.

 

Notes:

  • 1 <= grid.length = grid[0].length <= 50.
  • 0 <= grid[i][j] <= 1.



For each 1 in the grid, we paint all connected 1 with the next available color (2, 3, and so on). We also remember the size of the island we just painted with that color.

Then, we analyze all 0 in the grid, and sum sizes of connected islands (based on the island color). Note that the same island can connect to 0 more than once. The example below demonstrates this idea (the answer is highlighted):


class Solution {
public:
    int largestIsland(vector<vector<int>>& grid) 
    {
        vector<int> sizes={0,0};// sentinel values; colors start from 2 .
        for(auto i=0;i<grid.size();i++)
        {
            for(auto j=0;j<grid[i].size();j++)
            {
                if(grid[i][j]==1)
                {
                    sizes.push_back(paint(i,j,sizes.size(),grid));
                }
            }
        }
        
        for(auto i=0;i<grid.size();i++)
        {
            for(auto j=0;j<grid[i].size();j++)
            {
                if(grid[i][j]==0)
                {
                    unordered_set<int> s={get(i+1,j,grid),get(i-1,j,grid),get(i,j+1,grid),get(i,j-1,grid)};
                    res=max(res,1+accumulate(begin(s),end(s),0,[&](int a,int b){return a+sizes[b];}));
                }
            }
        }
        return res==0?grid.size()*grid[0].size():res;
    }
    
    int get(int i,int j,vector<vector<int>>& grid)
    {
        return (i<0 || j<0 || i>=grid.size() || j>=grid[0].size())?0:grid[i][j];
    }
    
    int paint(int i,int j,int clr,vector<vector<int>>& grid)
    {
        if(get(i,j,grid)!=1)
            return 0;
        grid[i][j]=clr;
        return 1+paint(i+1,j,clr,grid)+paint(i-1,j,clr,grid)+paint(i,j+1,clr,grid)+paint(i,j-1,clr,grid);
    }
    
private:
    int res=0;
};

猜你喜欢

转载自blog.csdn.net/xiaocong1990/article/details/80627946