题目描述
给你一个由 ‘1’(陆地)和 ‘0’(水)组成的的二维网格,请你计算网格中岛屿的数量。
岛屿总是被水包围,并且每座岛屿只能由水平方向或竖直方向上相邻的陆地连接形成。
此外,你可以假设该网格的四条边均被水包围。
示例 1:
输入:
[
['1','1','1','1','0'],
['1','1','0','1','0'],
['1','1','0','0','0'],
['0','0','0','0','0']
]
输出: 1
示例 2:
输入:
[
['1','1','0','0','0'],
['1','1','0','0','0'],
['0','0','1','0','0'],
['0','0','0','1','1']
]
输出: 3
解释: 每座岛屿只能由水平和/或竖直方向上相邻的陆地连接而成。
解答 By 海轰
提交代码(DFS)
void dfs(vector<vector<char>>& grid,int r,int c)
{
// 越界处理
if(!(0<=r&&r<grid.size()&&0<=c&&c<grid[0].size()))
return ;
// 不是陆地
if(grid[r][c]!='1')
return ;
grid[r][c]='2';// 将遍历过的地方更新为 防止循环
dfs(grid,r-1,c);
dfs(grid,r+1,c);
dfs(grid,r,c+1);
dfs(grid,r,c-1);
}
int numIslands(vector<vector<char>>& grid) {
int count=0;
for(int i=0;i<grid.size();++i)
{
for(int j=0;j<grid[0].size();++j)
{
if(grid[i][j]=='1')
{
dfs(grid,i,j);
++count;
}
}
}
return count;
}
运行结果
提交代码(BFS)
int numIslands(vector<vector<char>>& grid) {
int nr = grid.size();
if (!nr) return 0;
int nc = grid[0].size();
int num_islands = 0;
for (int r = 0; r < nr; ++r) {
for (int c = 0; c < nc; ++c) {
if (grid[r][c] == '1') {
++num_islands;
grid[r][c] = '0';
queue<pair<int, int>> neighbors;
neighbors.push({r, c});
while (!neighbors.empty()) {
auto rc = neighbors.front();
neighbors.pop();
int row = rc.first, col = rc.second;
if (row - 1 >= 0 && grid[row-1][col] == '1') {
neighbors.push({row-1, col});
grid[row-1][col] = '0';
}
if (row + 1 < nr && grid[row+1][col] == '1') {
neighbors.push({row+1, col});
grid[row+1][col] = '0';
}
if (col - 1 >= 0 && grid[row][col-1] == '1') {
neighbors.push({row, col-1});
grid[row][col-1] = '0';
}
if (col + 1 < nc && grid[row][col+1] == '1') {
neighbors.push({row, col+1});
grid[row][col+1] = '0';
}
}
}
}
}
return num_islands;
}
运行结果
解答
题目来源
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/number-of-islands