【并查集】93 岛屿数量

岛屿数量"/>

【并查集】93 岛屿数量

岛屿数量

• • 题解1 DFS（图论经典方法）
  • 题解2 BFS(遍历（DFS展开【顺序不同】）)
  • 题解3 并查集（学习理解）

岛屿总是被水包围，并且每座岛屿只能由水平方向和/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：

grid = [["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：

grid = [["1","1","0","0","0"],["1","1","0","0","0"],["0","0","1","0","0"],["0","0","0","1","1"]
]

输出：3

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 300
• grid[i][j] 的值为 ‘0’ 或 ‘1’

题解1 DFS（图论经典方法）

class Solution {int l, r;
public:void dfs(vector<vector<char>>& grid, int ro, int co){if(co < 0 || co >= r || ro < 0 || ro >= l ||  grid[ro][co] == '0')return;// 思想：把1附近相连（连通分量）置0// 相当于visited（面试时需要确定是否可以修改grid）grid[ro][co] = '0';dfs(grid, ro, co+1);dfs(grid, ro+1, co);dfs(grid, ro-1, co);dfs(grid, ro, co-1);} int numIslands(vector<vector<char>>& grid) {int ret = 0;l = grid.size();r = grid[0].size();for(int i = 0; i < l; i++){for(int j = 0; j < r; j++){if(grid[i][j] == '1'){ret += 1;dfs(grid, i, j);}}}return ret;}
};

题解2 BFS(遍历（DFS展开【顺序不同】）)

class Solution {
public:int numIslands(vector<vector<char>>& grid) {int ret = 0;int m = grid.size();int n = grid[0].size();for(int i = 0; i < m; i++){for(int j = 0; j < n; j++){if(grid[i][j] == '1'){ret += 1;grid[i][j] = '0';// BFS典型ADTqueue<pair<int, int>> nei;nei.push(make_pair(i, j));while(! nei.empty()){auto rc = nei.front();nei.pop();int r = rc.first;int c = rc.second;if(r >= 1 && grid[r-1][c] == '1'){nei.push(make_pair(r-1, c));grid[r-1][c] = '0';}if(r+1 < m && grid[r+1][c] == '1'){nei.push(make_pair(r+1, c));grid[r+1][c] = '0';}if(c >= 1 && grid[r][c-1] == '1'){nei.push(make_pair(r, c-1));grid[r][c-1] = '0';}if(c+1 < n && grid[r][c+1] == '1'){nei.push(make_pair(r, c+1));grid[r][c+1] = '0';}}}}}return ret;}
};

题解3 并查集（学习理解）

class UnionFind {vector<int> parent;vector<int> rank;int count;
public:UnionFind(vector<vector<char>>& grid) {count = 0;int m = grid.size();int n = grid[0].size();for (int i = 0; i < m; ++i) {for (int j = 0; j < n; ++j) {if (grid[i][j] == '1') {parent.push_back(i * n + j);++count;}else {parent.push_back(-1);}rank.push_back(0);}}}int find(int i) {if (parent[i] != i) {parent[i] = find(parent[i]);}return parent[i];}void unite(int x, int y) {int rootx = find(x);int rooty = find(y);if (rootx != rooty) {if (rank[rootx] < rank[rooty]) {swap(rootx, rooty);}parent[rooty] = rootx;if (rank[rootx] == rank[rooty]) rank[rootx] += 1;--count;}}int getCount() const {return count;}
};class Solution {
public:int numIslands(vector<vector<char>>& grid) {int nr = grid.size();if (!nr) return 0;int nc = grid[0].size();UnionFind uf(grid);int num_islands = 0;for (int r = 0; r < nr; ++r) {for (int c = 0; c < nc; ++c) {if (grid[r][c] == '1') {grid[r][c] = '0';if (r - 1 >= 0 && grid[r-1][c] == '1') uf.unite(r * nc + c, (r-1) * nc + c);if (r + 1 < nr && grid[r+1][c] == '1') uf.unite(r * nc + c, (r+1) * nc + c);if (c - 1 >= 0 && grid[r][c-1] == '1') uf.unite(r * nc + c, r * nc + c - 1);if (c + 1 < nc && grid[r][c+1] == '1') uf.unite(r * nc + c, r * nc + c + 1);}}}return uf.getCount();}
};