题目描述:
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.
要完成的函数:
int maxAreaOfIsland(vector<vector<int>>& grid)
说明:
1、给定一个二维矩阵,其中只含有0和1,0表示水域,1表示陆地,要求返回这片地方中最大的一块陆地的面积。
2、这其实是一道深度优先搜索或者广度优先搜索的题目。
由于DFS要用到递归,比较麻烦,所以笔者选择了BFS来实现,定义了一个队列。
代码如下:(附详解)
int maxAreaOfIsland(vector<vector<int>>& grid) { queue<int>q1; int hang=grid.size(),lie=grid[0].size(),sum=0,sum1=0,i=0,j=0,t1=0,t2=0; while(i<hang)//每一行的处理 { while(j<lie)//每一列的处理 { if(grid[i][j]==1)//如果搜索到一个1了 { q1.push(i);//把行坐标塞到队列里面去 q1.push(j);//把列坐标塞到队列里面去 grid[i][j]=0;//并将这个点改为陆地,避免后面再次搜索到,重复计算 sum1=0;//统计当前陆地的面积 while(!q1.empty())//当队列非空时,迭代处理 { t1=q1.front();//取出行坐标 q1.pop(); t2=q1.front();//取出列坐标 q1.pop(); sum1++; if(t1-1>=0&&grid[t1-1][t2]==1)//判断上方有没有陆地 { q1.push(t1-1); q1.push(t2); grid[t1-1][t2]=0;//置为0,避免再次搜索到,重复计算 } if(t1+1<hang&&grid[t1+1][t2]==1)//判断下方有没有陆地 { q1.push(t1+1); q1.push(t2); grid[t1+1][t2]=0; } if(t2-1>=0&&grid[t1][t2-1]==1)//判断左方有没有陆地 { q1.push(t1); q1.push(t2-1); grid[t1][t2-1]=0; } if(t2+1<lie&&grid[t1][t2+1]==1)//判断右方有没有陆地 { q1.push(t1); q1.push(t2+1); grid[t1][t2+1]=0; } } sum=max(sum,sum1);//取每次陆地面积的最大值 } j++; } i++; j=0;//j=0记得要加上 } return sum; }
上述代码实测30ms,beats 80.17% of cpp submissions。
原文地址:https://www.cnblogs.com/king-3/p/9188490.html