LeetCode200——Number of Islands

最新推荐文章于 2021-12-21 14:26:15 发布

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本文介绍了一种算法，用于计算二维网格地图中由'1'（陆地）和'0'（水域）组成的岛屿数量。通过遍历每个网格并使用深度优先搜索的方法标记相连的陆地，实现了准确计算岛屿数目。

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Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

Example 1:

Answer: 1

Example 2:

Answer: 3

实现：

class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
int count = 0;
for (int i = 0; i < grid.size(); i++) {
for (int j = 0; j < grid[i].size(); j++) {
if (grid[i][j] == '1') {
count++;
colorVec(grid, i, j);
}
}
}
return count;
}

int colorVec(vector<vector<char>>& grid, int i, int j) {
vector<Point> stack;
stack.push_back(Point(i, j));
while (!stack.empty()) {
Point p = stack.back();
stack.pop_back();
int x = p.x;
int y = p.y;
grid[x][y] = '2';
if (x-1 >= 0 && grid[x-1][y] == '1') {
stack.push_back(Point(x-1, y));
}
if (x+1 < grid.size() && grid[x+1][y] == '1') {
stack.push_back(Point(x+1, y));
}
if (y+1 < grid[x].size() && grid[x][y+1] == '1') {
stack.push_back(Point(x, y+1));
}
if (y-1 >= 0 && grid[x][y-1] == '1') {
stack.push_back(Point(x, y-1));
}
}
}
};