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| 1 | +# LeetCode 图解 | |
| 2 | + |
| 3 | +> 本文首发于公众号「图解面试算法」,是 [图解 LeetCode](<https://github.com/MisterBooo/LeetCodeAnimation>) 系列文章之一。 |
| 4 | +> |
| 5 | +> 同步博客:https://www.algomooc.com |
| 6 | + |
| 7 | +本题解作者:nettee |
| 8 | + |
| 9 | +## 题目描述 |
| 10 | + |
| 11 | +给定一个包含了一些 `0` 和 `1` 的非空二维数组 `grid`。 |
| 12 | + |
| 13 | +一个**岛屿**是由一些相邻的 `1` (代表土地) 构成的组合,这里的「相邻」要求两个 `1` 必须在水平或者竖直方向上相邻。你可以假设 `grid` 的四个边缘都被 `0`(代表水)包围着。 |
| 14 | + |
| 15 | +找到给定的二维数组中最大的岛屿面积。(如果没有岛屿,则返回面积为 `0`。) |
| 16 | + |
| 17 | +**示例 1:** |
| 18 | + |
| 19 | +``` |
| 20 | +[[0,0,1,0,0,0,0,1,0,0,0,0,0], |
| 21 | + [0,0,0,0,0,0,0,1,1,1,0,0,0], |
| 22 | + [0,1,1,0,1,0,0,0,0,0,0,0,0], |
| 23 | + [0,1,0,0,1,1,0,0,1,0,1,0,0], |
| 24 | + [0,1,0,0,1,1,0,0,1,1,1,0,0], |
| 25 | + [0,0,0,0,0,0,0,0,0,0,1,0,0], |
| 26 | + [0,0,0,0,0,0,0,1,1,1,0,0,0], |
| 27 | + [0,0,0,0,0,0,0,1,1,0,0,0,0]] |
| 28 | +``` |
| 29 | + |
| 30 | + |
| 31 | +对于上面这个给定矩阵应返回 6。注意答案不应该是 11 ,因为岛屿只能包含水平或垂直的四个方向的 `1` 。 |
| 32 | + |
| 33 | +**示例 2:** |
| 34 | + |
| 35 | +``` |
| 36 | +[[0,0,0,0,0,0,0,0]] |
| 37 | +``` |
| 38 | + |
| 39 | + |
| 40 | +对于上面这个给定的矩阵, 返回 0。 |
| 41 | + |
| 42 | +注意: 给定的矩阵 `grid` 的长度和宽度都不超过 50。 |
| 43 | + |
| 44 | +## 题目解析 |
| 45 | + |
| 46 | +这道题的主要思路是深度优先搜索。每次走到一个是 1 的格子,就搜索整个岛屿,并计算当前岛屿的面积。最后返回岛屿面积的最大值。 |
| 47 | + |
| 48 | +网格可以看成是一个无向图的结构,每个格子和它上下左右的四个格子相邻。如果四个相邻的格子坐标合法,且是陆地,就可以继续搜索。 |
| 49 | + |
| 50 | +在深度优先搜索的时候要注意避免重复遍历。我们可以把已经遍历过的陆地改成 2,这样遇到 2 我们就知道已经遍历过这个格子了,不进行重复遍历。 |
| 51 | + |
| 52 | +## 动画理解 |
| 53 | + |
| 54 | + |
| 55 | + |
| 56 | +## 参考代码 |
| 57 | + |
| 58 | +C++ 代码: |
| 59 | + |
| 60 | +```C++ |
| 61 | +class Solution { |
| 62 | +public: |
| 63 | + int maxAreaOfIsland(vector<vector<int>>& grid) { |
| 64 | + int res = 0; |
| 65 | + for (int r = 0; r < grid.size(); r++) { |
| 66 | + for (int c = 0; c < grid[0].size(); c++) { |
| 67 | + if (grid[r][c] == 1) { |
| 68 | + int a = area(grid, r, c); |
| 69 | + res = max(res, a); |
| 70 | + } |
| 71 | + } |
| 72 | + } |
| 73 | + return res; |
| 74 | + } |
| 75 | + |
| 76 | + int area(vector<vector<int>>& grid, int r, int c) { |
| 77 | + if (!(inArea(grid, r, c))) { |
| 78 | + return 0; |
| 79 | + } |
| 80 | + if (grid[r][c] != 1) { |
| 81 | + return 0; |
| 82 | + } |
| 83 | + grid[r][c] = 2; |
| 84 | + |
| 85 | + return 1 |
| 86 | + + area(grid, r - 1, c) |
| 87 | + + area(grid, r + 1, c) |
| 88 | + + area(grid, r, c - 1) |
| 89 | + + area(grid, r, c + 1); |
| 90 | + } |
| 91 | + |
| 92 | + bool inArea(vector<vector<int>>& grid, int r, int c) { |
| 93 | + return 0 <= r && r < grid.size() |
| 94 | + && 0 <= c && c < grid[0].size(); |
| 95 | + } |
| 96 | +}; |
| 97 | +``` |
| 98 | + |
| 99 | +Java 代码: |
| 100 | + |
| 101 | +```Java |
| 102 | +class Solution { |
| 103 | + public int maxAreaOfIsland(int[][] grid) { |
| 104 | + int res = 0; |
| 105 | + for (int r = 0; r < grid.length; r++) { |
| 106 | + for (int c = 0; c < grid[0].length; c++) { |
| 107 | + if (grid[r][c] == 1) { |
| 108 | + int a = area(grid, r, c); |
| 109 | + res = Math.max(res, a); |
| 110 | + } |
| 111 | + } |
| 112 | + } |
| 113 | + return res; |
| 114 | + } |
| 115 | + |
| 116 | + int area(int[][] grid, int r, int c) { |
| 117 | + if (!inArea(grid, r, c)) { |
| 118 | + return 0; |
| 119 | + } |
| 120 | + if (grid[r][c] != 1) { |
| 121 | + return 0; |
| 122 | + } |
| 123 | + grid[r][c] = 2; |
| 124 | + |
| 125 | + return 1 |
| 126 | + + area(grid, r - 1, c) |
| 127 | + + area(grid, r + 1, c) |
| 128 | + + area(grid, r, c - 1) |
| 129 | + + area(grid, r, c + 1); |
| 130 | + } |
| 131 | + |
| 132 | + boolean inArea(int[][] grid, int r, int c) { |
| 133 | + return 0 <= r && r < grid.length |
| 134 | + && 0 <= c && c < grid[0].length; |
| 135 | + } |
| 136 | +} |
| 137 | +``` |
| 138 | + |
| 139 | +Python 代码: |
| 140 | + |
| 141 | +```Python |
| 142 | +class Solution: |
| 143 | + def maxAreaOfIsland(self, grid: List[List[int]]) -> int: |
| 144 | + res = 0 |
| 145 | + for r in range(len(grid)): |
| 146 | + for c in range(len(grid[0])): |
| 147 | + if grid[r][c] == 1: |
| 148 | + a = self.area(grid, r, c) |
| 149 | + res = max(res, a) |
| 150 | + return res |
| 151 | + |
| 152 | + def area(self, grid: List[List[int]], r: int, c: int) -> int: |
| 153 | + if not self.inArea(grid, r, c): |
| 154 | + return 0 |
| 155 | + if grid[r][c] != 1: |
| 156 | + return 0 |
| 157 | + grid[r][c] = 2 |
| 158 | + |
| 159 | + return 1 \ |
| 160 | + + self.area(grid, r - 1, c) \ |
| 161 | + + self.area(grid, r + 1, c) \ |
| 162 | + + self.area(grid, r, c - 1) \ |
| 163 | + + self.area(grid, r, c + 1) |
| 164 | + |
| 165 | + def inArea(self, grid: List[List[int]], r: int, c: int) -> bool: |
| 166 | + return 0 <= r < len(grid) and 0 <= c < len(grid[0]) |
| 167 | +``` |
| 168 | + |
| 169 | + |
| 170 | + |
| 171 | +## 复杂度分析 |
| 172 | + |
| 173 | +设网格的边长为 n,则时间复杂度为 O(n2)。 |
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