|
| 1 | +# 岛屿数量 |
| 2 | + |
| 3 | +> 难度：中等 |
| 4 | +> |
| 5 | +> https://leetcode-cn.com/problems/number-of-islands/ |
| 6 | +
|
| 7 | +## 题目 |
| 8 | + |
| 9 | +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 |
| 10 | +量。 |
| 11 | + |
| 12 | +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。 |
| 13 | + |
| 14 | +此外，你可以假设该网格的四条边均被水包围。 |
| 15 | + |
| 16 | +### 示例 |
| 17 | + |
| 18 | +#### 示例 1： |
| 19 | + |
| 20 | +``` |
| 21 | +输入：grid = [ |
| 22 | + ["1","1","1","1","0"], |
| 23 | + ["1","1","0","1","0"], |
| 24 | + ["1","1","0","0","0"], |
| 25 | + ["0","0","0","0","0"] |
| 26 | +] |
| 27 | +输出：1 |
| 28 | +``` |
| 29 | + |
| 30 | +#### 示例 2： |
| 31 | + |
| 32 | +``` |
| 33 | +输入：grid = [ |
| 34 | + ["1","1","0","0","0"], |
| 35 | + ["1","1","0","0","0"], |
| 36 | + ["0","0","1","0","0"], |
| 37 | + ["0","0","0","1","1"] |
| 38 | +] |
| 39 | +输出：3 |
| 40 | +``` |
| 41 | + |
| 42 | +## 解题 |
| 43 | + |
| 44 | +### 深度优先遍历 |
| 45 | + |
| 46 | +```typescript |
| 47 | +/** |
| 48 | + * 深度优先遍历 |
| 49 | + * @desc 时间复杂度 O(NM) 空间复杂度 O(NM) |
| 50 | + * @param grid |
| 51 | + * @returns |
| 52 | + */ |
| 53 | +export function numIslands(grid: string[][]): number { |
| 54 | + if (grid.length === 0 || grid[0].length === 0) return 0; |
| 55 | + |
| 56 | + const m = grid.length; |
| 57 | + const n = grid[0].length; |
| 58 | + let result = 0; |
| 59 | + |
| 60 | + for (let i = 0; i < m; i++) { |
| 61 | + for (let j = 0; j < n; j++) { |
| 62 | + if (grid[i][j] !== '1') continue; |
| 63 | + result++; |
| 64 | + dfs(grid, m, n, i, j); |
| 65 | + } |
| 66 | + } |
| 67 | + |
| 68 | + return result; |
| 69 | + |
| 70 | + function dfs( |
| 71 | + grid: string[][], |
| 72 | + m: number, |
| 73 | + n: number, |
| 74 | + i: number, |
| 75 | + j: number |
| 76 | + ): void { |
| 77 | + if (i < 0 || j < 0 || i >= m || j >= n || grid[i][j] === '0') return; |
| 78 | + |
| 79 | + grid[i][j] = '0'; |
| 80 | + dfs(grid, m, n, i + 1, j); |
| 81 | + dfs(grid, m, n, i - 1, j); |
| 82 | + dfs(grid, m, n, i, j + 1); |
| 83 | + dfs(grid, m, n, i, j - 1); |
| 84 | + } |
| 85 | +} |
| 86 | +``` |
| 87 | + |
| 88 | +### 广度优先遍历 |
| 89 | + |
| 90 | +```typescript |
| 91 | +/** |
| 92 | + * 广度优先遍历 |
| 93 | + * @desc 时间复杂度 O(NM) 空间复杂度 O(NM) |
| 94 | + * @param grid |
| 95 | + * @returns |
| 96 | + */ |
| 97 | +export function numIslands2(grid: string[][]): number { |
| 98 | + if (grid.length === 0 || grid[0].length === 0) return 0; |
| 99 | + |
| 100 | + const m = grid.length; |
| 101 | + const n = grid[0].length; |
| 102 | + let result = 0; |
| 103 | + |
| 104 | + // 生成队列key |
| 105 | + const generateKey = (i: number, j: number): number => |
| 106 | + i * n + j; /* j 恒小于 n */ |
| 107 | + |
| 108 | + for (let i = 0; i < m; i++) { |
| 109 | + for (let j = 0; j < n; j++) { |
| 110 | + if (grid[i][j] !== '1') continue; |
| 111 | + result++; |
| 112 | + grid[i][j] = '0'; |
| 113 | + |
| 114 | + const queue = [generateKey(i, j)]; |
| 115 | + |
| 116 | + while (queue.length) { |
| 117 | + const key = queue.pop()!; |
| 118 | + const row = (key / n) >> 0; |
| 119 | + const col = key % n; |
| 120 | + console.log(row, col); |
| 121 | + if (row - 1 >= 0 && grid[row - 1][col] === '1') { |
| 122 | + queue.unshift(generateKey(row - 1, col)); |
| 123 | + grid[row - 1][col] = '0'; |
| 124 | + } |
| 125 | + if (row + 1 < m && grid[row + 1][col] === '1') { |
| 126 | + queue.unshift(generateKey(row + 1, col)); |
| 127 | + grid[row + 1][col] = '0'; |
| 128 | + } |
| 129 | + if (col - 1 >= 0 && grid[row][col - 1] === '1') { |
| 130 | + queue.unshift(generateKey(row, col - 1)); |
| 131 | + grid[row][col - 1] = '0'; |
| 132 | + } |
| 133 | + if (col + 1 < n && grid[row][col + 1] === '1') { |
| 134 | + queue.unshift(generateKey(row, col + 1)); |
| 135 | + grid[row][col + 1] = '0'; |
| 136 | + } |
| 137 | + } |
| 138 | + } |
| 139 | + } |
| 140 | + |
| 141 | + return result; |
| 142 | +} |
| 143 | +``` |
| 144 | + |
| 145 | +### 并查集 |
| 146 | + |
| 147 | +```typescript |
| 148 | +/** |
| 149 | + * 并查集 |
| 150 | + * @desc 时间复杂度 O(NM) 空间复杂度 O(NM) |
| 151 | + * @param grid |
| 152 | + * @returns |
| 153 | + */ |
| 154 | +export function numIslands3(grid: string[][]): number { |
| 155 | + if (grid.length === 0 || grid[0].length === 0) return 0; |
| 156 | + |
| 157 | + const unionFind = new UnionFind(grid); |
| 158 | + const { m, n } = unionFind; |
| 159 | + |
| 160 | + for (let i = 0; i < m; i++) { |
| 161 | + for (let j = 0; j < n; j++) { |
| 162 | + if (grid[i][j] === '1') { |
| 163 | + grid[i][j] = '0'; |
| 164 | + if (i - 1 >= 0 && grid[i - 1][j] === '1') { |
| 165 | + unionFind.union( |
| 166 | + unionFind.generateKey(i, j), |
| 167 | + unionFind.generateKey(i - 1, j) |
| 168 | + ); |
| 169 | + } |
| 170 | + if (i + 1 < m && grid[i + 1][j] === '1') { |
| 171 | + unionFind.union( |
| 172 | + unionFind.generateKey(i, j), |
| 173 | + unionFind.generateKey(i + 1, j) |
| 174 | + ); |
| 175 | + } |
| 176 | + if (j - 1 >= 0 && grid[i][j - 1] === '1') { |
| 177 | + unionFind.union( |
| 178 | + unionFind.generateKey(i, j), |
| 179 | + unionFind.generateKey(i, j - 1) |
| 180 | + ); |
| 181 | + } |
| 182 | + if (j + 1 < n && grid[i][j + 1] === '1') { |
| 183 | + unionFind.union( |
| 184 | + unionFind.generateKey(i, j), |
| 185 | + unionFind.generateKey(i, j + 1) |
| 186 | + ); |
| 187 | + } |
| 188 | + } |
| 189 | + } |
| 190 | + } |
| 191 | + |
| 192 | + return unionFind.count; |
| 193 | +} |
| 194 | + |
| 195 | +class UnionFind { |
| 196 | + count: number; // 记录item为1的个数 |
| 197 | + parent: number[]; // index为每个item的唯一key，value为它们合并后的根位置 |
| 198 | + rank: number[]; // 记录每个根位置的合并数量 |
| 199 | + m: number; |
| 200 | + n: number; |
| 201 | + |
| 202 | + constructor(grid: string[][]) { |
| 203 | + // 初始化 |
| 204 | + this.count = 0; |
| 205 | + const m = (this.m = grid.length); |
| 206 | + const n = (this.n = grid[0].length); |
| 207 | + this.parent = new Array(m * n).fill(-1); |
| 208 | + this.rank = new Array(m * n).fill(0); |
| 209 | + |
| 210 | + for (let i = 0; i < m; i++) { |
| 211 | + for (let j = 0; j < n; j++) { |
| 212 | + const key = this.generateKey(i, j); |
| 213 | + if (grid[i][j] === '1') { |
| 214 | + this.parent[key] = key; |
| 215 | + this.count++; |
| 216 | + } |
| 217 | + } |
| 218 | + } |
| 219 | + } |
| 220 | + |
| 221 | + /** |
| 222 | + * 生成唯一keys |
| 223 | + * @param i |
| 224 | + * @param j |
| 225 | + * @returns |
| 226 | + */ |
| 227 | + generateKey(i: number, j: number): number { |
| 228 | + return i * this.n + j; |
| 229 | + } |
| 230 | + |
| 231 | + find(i: number): number { |
| 232 | + if (this.parent[i] !== i) this.parent[i] = this.find(this.parent[i]); |
| 233 | + return this.parent[i]; |
| 234 | + } |
| 235 | + |
| 236 | + /** |
| 237 | + * 合并操作 |
| 238 | + * @param x |
| 239 | + * @param y |
| 240 | + */ |
| 241 | + union(x: number, y: number) { |
| 242 | + // 找到x,y的根位置 |
| 243 | + // 这里可以确保x,y的item都为1 |
| 244 | + const xRoot = this.find(x); |
| 245 | + const yRoot = this.find(y); |
| 246 | + |
| 247 | + // xRoot === yRoot 代表已经合并了 |
| 248 | + if (xRoot === yRoot) return; |
| 249 | + |
| 250 | + // 合并操作 |
| 251 | + if (this.rank[xRoot] > this.rank[yRoot]) { |
| 252 | + this.parent[yRoot] = xRoot; |
| 253 | + } else if (this.rank[xRoot] < this.rank[yRoot]) { |
| 254 | + this.parent[xRoot] = yRoot; |
| 255 | + } else { |
| 256 | + this.parent[yRoot] = xRoot; |
| 257 | + this.rank[xRoot] += 1; |
| 258 | + } |
| 259 | + this.count--; |
| 260 | + } |
| 261 | +} |
| 262 | +``` |
0 commit comments