[LeetCode] 463. Island Perimeter 岛的周长
2023-09-27 14:28:37 时间
You are given a map in form of a two-dimensional integer grid where 1 represents land and 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn't have "lakes" (water inside that isn't connected to the water around the island). One cell is a square with side length 1. The grid is rectangular, width and height don't exceed 100. Determine the perimeter of the island.
Example:
[[0,1,0,0], [1,1,1,0], [0,1,0,0], [1,1,0,0]] Answer: 16 Explanation: The perimeter is the 16 yellow stripes in the image below:
给一个二维的格子图,格子是1表示陆地,0表示水,格子水平或者垂直连接,若干的格子连在一起形成一个小岛,只有一个相连的岛,且岛中没有湖,求岛的周长。
解法1: 遇到为1的格子,检验四个方向是否和陆地相连
解法2: 遇到为1的格子,检查上面和左面是否和陆地相连
Java:
public class Solution { public int islandPerimeter(int[][] grid) { int islands = 0, neighbours = 0; for (int i = 0; i < grid.length; i++) { for (int j = 0; j < grid[i].length; j++) { if (grid[i][j] == 1) { islands++; // count islands if (i < grid.length - 1 && grid[i + 1][j] == 1) neighbours++; // count down neighbours if (j < grid[i].length - 1 && grid[i][j + 1] == 1) neighbours++; // count right neighbours } } } return islands * 4 - neighbours * 2; } }
Python:
def islandPerimeter(self, grid): """ :type grid: List[List[int]] :rtype: int """ ans = 0 for i in range(len(grid)): for j in range(len(grid[0])): if grid[i][j] == 1: ans += 4 if i > 0 and grid[i-1][j] == 1: ans -= 2 if j > 0 and grid[i][j-1] == 1: ans -= 2 return ans
Python:
class Solution(object): def islandPerimeter(self, grid): """ :type grid: List[List[int]] :rtype: int """ if not grid: return 0 def sum_adjacent(i, j): adjacent = (i + 1, j), (i - 1, j), (i, j + 1), (i, j - 1), res = 0 for x, y in adjacent: if x < 0 or y < 0 or x == len(grid) or y == len(grid[0]) or grid[x][y] == 0: res += 1 return res count = 0 for i in range(len(grid)): for j in range(len(grid[0])): if grid[i][j] == 1: count += sum_adjacent(i, j) return count
Python:
class Solution: def islandPerimeter(self, grid): if not grid: return 0 N = len(grid) M = len(grid[0]) ans = 0 for i in range(N): for j in range(M): for k in [(-1, 0), (0, 1), (1, 0), (0, -1)]: if i+k[0] < 0 or i + k[0] >= N or j+k[1] < 0 or j+k[1] >= M or grid[i+k[0]][j+k[1]] == 0: ans += 1 return ans
C++:
class Solution { public: int islandPerimeter(vector<vector<int>>& grid) { if (grid.empty() || grid[0].empty()) return 0; int m = grid.size(), n = grid[0].size(), res = 0; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (grid[i][j] == 0) continue; if (j == 0 || grid[i][j - 1] == 0) ++res; if (i == 0 || grid[i - 1][j] == 0) ++res; if (j == n - 1 || grid[i][j + 1] == 0) ++res; if (i == m - 1 || grid[i + 1][j] == 0) ++res; } } return res; } };
C++:
class Solution { public: int islandPerimeter(vector<vector<int>>& grid) { if (grid.empty() || grid[0].empty()) return 0; int res = 0, m = grid.size(), n = grid[0].size(); for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (grid[i][j] == 0) continue; res += 4; if (i > 0 && grid[i - 1][j] == 1) res -= 2; if (j > 0 && grid[i][j - 1] == 1) res -= 2; } } return res; } };
Followup: 如果不止一个岛屿
All LeetCode Questions List 题目汇总
相关文章
- Leetcode: Shortest Way to Form String
- Leetcode: Poor Pigs
- Leetcode: Convert a Number to Hexadecimal
- LeetCode --- 58. Length of Last Word
- 【Leetcode】746. 使用最小花费爬楼梯(简单)
- [LeetCode]剑指 Offer 21. 调整数组顺序使奇数位于偶数前面
- 181、【动态规划】leetcode ——72. 编辑距离(C++版本)
- 【LeetCode】203. Remove Linked List Elements
- [leetcode]Permutation Sequence
- [LeetCode] Student Attendance Record I 学生出勤记录之一
- [LeetCode] Bulls and Cows 公母牛游戏