给定一个包含了一些 0 和 1的非空二维数组 grid
, 一个 岛屿 是由四个方向 (水平或垂直) 的 1
(代表土地) 构成的组合。你可以假设二维矩阵的四个边缘都被水包围着。
找到给定的二维数组中最大的岛屿面积。(如果没有岛屿,则返回面积为0。)
示例 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]]
对于上面这个给定矩阵应返回 6
。注意答案不应该是11,因为岛屿只能包含水平或垂直的四个方向的‘1’。
示例 2:
[[0,0,0,0,0,0,0,0]]
对于上面这个给定的矩阵, 返回 0
。
注意: 给定的矩阵grid
的长度和宽度都不超过 50。
class Solution {
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
int area=0;
int temp=0;
for(int i=0;i<grid.size();i++){
for(int j=0;j<grid[0].size();j++){
if(grid[i][j]==1){
temp=calcMax(grid,i,j);
area=max(area,temp);
}
}
}
return area;
}
int calcMax(vector<vector<int>>& grid,int i,int j){
int n = 1;
grid[i][j] = 0;
if(i+1<grid.size() && grid[i+1][j] == 1){
n += calcMax(grid, i+1, j);
}
if(j+1<grid[0].size() && grid[i][j+1] == 1){
n += calcMax(grid, i, j+1);
}
if(i-1>=0 && grid[i-1][j] == 1){
n += calcMax(grid, i-1, j);
}
if(j-1>=0 && grid[i][j-1] == 1){
n += calcMax(grid, i, j-1);
}
return n;
}
};
此方法为递归调用,检测为1的上下左右,并把1置0
class Solution {
public:
int maxAreaOfIsland(vector<vector<int>>& grid) {
int m=grid.size(), n=grid[0].size();
int max_a=0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j]==1){
int a = islandDFS(grid, m, n, i, j);
max_a = max(max_a, a);
}
}
}
return max_a;
}
int islandDFS(vector<vector<int>>& grid, int& m, int& n, int i, int j){
if (i<0 || i>=m || j<0 || j>=n || grid[i][j]==0)
return 0;
else{
grid[i][j] = 0;
return 1+islandDFS(grid, m, n, i+1, j)+islandDFS(grid, m, n, i-1, j)+
islandDFS(grid, m, n, i, j+1)+islandDFS(grid, m, n, i, j-1);
}
}
};