并查集）

DFS解法：

 1 class Solution {
 2 public:
 3     typedef pair<int,int> pii;
 4     int orientation[8] = {1, 0, -1, 0, 0, 1, 0, -1};
 5     int n,m;
 6     vector<vector<char>> grid;
 7     map<pii, bool> visited;
 8     void dfs(pii start) {
 9         visited[start] = true;
10         for (int i = 0; i < 4; ++i) {
11             int x = start.first + orientation[i * 2], y = start.second + orientation[i * 2 + 1];
12             if (x >= 0 && x < n && y >= 0 && y < m) { //注意边界
13                 pii temp = make_pair(x, y);
14                 if (!visited[temp] && grid[x][y] == '1') {
15                     dfs(temp);
16                 }
17             }
18 
19         }
20     }
21     int numIslands(vector<vector<char>>& grid) {
22         this -> grid = grid;
23         n = grid.size();
24         m = grid[0].size();
25         int result = 0;
26         for (int i = 0; i < n; ++i) {
27             for (int j = 0; j < m; ++j) {
28                 pii temp = make_pair(i, j);
29                 if (grid[i][j] == '1' && !visited[temp]) {
30                     result ++;
31                     dfs(temp);
32                 }
33             }
34         }
35         return result;
36     }
37 };

并查集：

 1 class Solution {
 2 public:
 3     typedef pair<int, int> pii;
 4     map<pii, pii> parent;
 5     int orientation[4] = {1, 0, 0, 1}; //已经往右和下遍历了，检查联通性时不需要判定左和上
 6     pii find(pii x) {
 7         if (parent[x] == x) {
 8             return x;
 9         } else {
10             return parent[x] = find(parent[x]);
11         }
12     } 
13     void merge(pii x, pii y) {
14         pii xx = find(x), yy = find(y);
15         if (xx != yy) {
16             parent[yy] = xx;
17         }
18     }
19     int numIslands(vector<vector<char>>& grid) {
20         int result = 0;
21         for (int i = 0; i < grid.size(); ++i) {
22             for (int j = 0; j < grid[0].size(); ++j) {
23                 if (grid[i][j] == '1') {
24                     parent[make_pair(i, j)] = make_pair(i, j);
25                 }
26             }
27         }
28         for (int i = 0; i < grid.size(); ++i) {
29             for (int j = 0; j < grid[0].size(); ++j) {
30                 if (grid[i][j] == '1') {
31                     for (int k = 0; k < 2; ++k) {
32                         int x = i + orientation[k * 2], y = j + orientation[k * 2 + 1];
33                         if (x >= 0 && x < grid.size() && y >= 0 && y < grid[0].size()) {
34                             if (grid[x][y] == '1') {
35                                 merge(make_pair(i, j), make_pair(x, y));
36                             }
37                         }
38                     }
39                 }
40             }
41         }
42         for (int i = 0; i < grid.size(); ++i) {
43             for (int j = 0; j < grid[0].size(); ++j) {
44                 pii temp = make_pair(i, j);
45                 if (parent.count(temp) > 0 && parent[temp] == temp) {
46                     result++;
47                 }
48             }
49         }
50         return result;
51     }
52 };

posted on 2025-09-01 23:38 Coder何阅读(8) 评论(0) 收藏举报