305 Number of Islands II
A 2d grid map of m rows and n columns is initially filled with water. We may perform an addLand operation which turns the water at position (row, col) into a land. Given a list of positions to operate, count the number of islands after each addLand operation. 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:
Given m = 3, n = 3, positions = [[0,0], [0,1], [1,2], [2,1]].
Initially, the 2d grid grid is filled with water. (Assume 0 represents water and 1 represents land).
0 0 0
0 0 0
0 0 0
Operation #1: addLand(0, 0) turns the water at grid[0][0] into a land.
1 0 0
0 0 0 Number of islands = 1
0 0 0
Operation #2: addLand(0, 1) turns the water at grid[0][1] into a land.
1 1 0
0 0 0 Number of islands = 1
0 0 0
Operation #3: addLand(1, 2) turns the water at grid[1][2] into a land.
1 1 0
0 0 1 Number of islands = 2
0 0 0
Operation #4: addLand(2, 1) turns the water at grid[2][1] into a land.
1 1 0
0 0 1 Number of islands = 3
0 1 0
We return the result as an array: [1, 1, 2, 3]
Challenge:
Can you do it in time complexity O(k log mn), where k is the length of the positions?
分析& 代码
使用Union-Find方法。
应该还是比较标准的uf，但是开始自己看不出来。把二维压成一维，给一个编号，其他的基本上没有变化。
伪代码解释：
对于每一个新加的点：
cnt++;
//uf做的部分就是把连起来了的岛融合在一起，调整这个cnt的
把自己的parent标记成自己
对于每一个方向：
如果新的点不在范围内，或者这个点没有之前标记过（就是不存在需要融合的情况）
continue
找到这个新位置的root,如果这个newRoot和当前root不是一个，那么就说明需要融合
roots[root] = newRoot;
cnt--;
root = newRoot.//这句不可以少，因为如果一个新加的点让cnt+1，但是它成功地连起来了多个点，就会出现多减了的情况，所以要更新root
把cnt加入res中
返回res
代码：
static final int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
public List<Integer> numIslands2(int m, int n, int[][] positions) {
List<Integer> res = new ArrayList<Integer>();
if(m <= 0 || n <= 0 || positions.length == 0 || positions[0].length == 0) {
return res;
}
int[] roots = new int[m * n];
Arrays.fill(roots, -1);
int cnt = 0;
for(int[] point: positions) {
int root = point[0] * n + point[1];
roots[root] = root;
cnt++;
for(int[] dir: dirs) {
int newX = point[0] + dir[0];
int newY = point[1] + dir[1];
int newId = newX * n + newY;
if(newX < 0 || newY < 0 || newX >= m || newY >=n || roots[newId] == -1) {
continue;
}
int newRoot = find(roots, newId);
if(newRoot != root) {
roots[root] = newRoot;
root = newRoot;
cnt--;
}
}
res.add(cnt);
}
return res;
}
private int find(int[] roots, int root) {
if(roots[root] == root) {
return root;
}
return find(roots, roots[root]);
}