258 lines
8.5 KiB
Markdown
258 lines
8.5 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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# 463. 岛屿的周长
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[力扣题目链接](https://leetcode.cn/problems/island-perimeter/)
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## 思路
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岛屿问题最容易让人想到BFS或者DFS，但是这道题还真的没有必要，别把简单问题搞复杂了。
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### 解法一：
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遍历每一个空格，遇到岛屿，计算其上下左右的情况，遇到水域或者出界的情况，就可以计算边了。
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如图：
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<img src='https://code-thinking.cdn.bcebos.com/pics/463.岛屿的周长.png' width=600> </img></div>
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C++代码如下：（详细注释）
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```CPP
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class Solution {
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public:
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int direction[4][2] = {0, 1, 1, 0, -1, 0, 0, -1};
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int islandPerimeter(vector<vector<int>>& grid) {
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int result = 0;
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for (int i = 0; i < grid.size(); i++) {
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for (int j = 0; j < grid[0].size(); j++) {
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if (grid[i][j] == 1) {
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for (int k = 0; k < 4; k++) { // 上下左右四个方向
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int x = i + direction[k][0];
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int y = j + direction[k][1]; // 计算周边坐标x,y
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if (x < 0 // i在边界上
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|| x >= grid.size() // i在边界上
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|| y < 0 // j在边界上
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|| y >= grid[0].size() // j在边界上
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|| grid[x][y] == 0) { // x,y位置是水域
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result++;
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}
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}
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}
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}
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}
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return result;
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}
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};
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```
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### 解法二：
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计算出总的岛屿数量，因为有一对相邻两个陆地，边的总数就减2，那么在计算出相邻岛屿的数量就可以了。
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result = 岛屿数量 * 4 - cover * 2;
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如图：
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<img src='https://code-thinking.cdn.bcebos.com/pics/463.岛屿的周长1.png' width=600> </img></div>
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C++代码如下：（详细注释）
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```CPP
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class Solution {
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public:
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int islandPerimeter(vector<vector<int>>& grid) {
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int sum = 0; // 陆地数量
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int cover = 0; // 相邻数量
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for (int i = 0; i < grid.size(); i++) {
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for (int j = 0; j < grid[0].size(); j++) {
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if (grid[i][j] == 1) {
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sum++;
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// 统计上边相邻陆地
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if(i - 1 >= 0 && grid[i - 1][j] == 1) cover++;
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// 统计左边相邻陆地
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if(j - 1 >= 0 && grid[i][j - 1] == 1) cover++;
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// 为什么没统计下边和右边？ 因为避免重复计算
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}
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}
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}
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return sum * 4 - cover * 2;
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}
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};
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```
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## 其他语言版本
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Java：
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```java
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// 解法一
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class Solution {
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// 上下左右 4 个方向
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int[] dirx = {-1, 1, 0, 0};
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int[] diry = {0, 0, -1, 1};
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public int islandPerimeter(int[][] grid) {
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int m = grid.length;
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int n = grid[0].length;
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int res = 0; // 岛屿周长
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (grid[i][j] == 1) {
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for (int k = 0; k < 4; k++) {
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int x = i + dirx[k];
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int y = j + diry[k];
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// 当前位置是陆地，并且从当前位置4个方向扩展的“新位置”是“水域”或“新位置“越界，则会为周长贡献一条边
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if (x < 0 || x >= m || y < 0 || y >= n || grid[x][y] == 0) {
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res++;
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continue;
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}
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}
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}
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}
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}
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return res;
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}
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}
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// 解法二
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class Solution {
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public int islandPerimeter(int[][] grid) {
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// 计算岛屿的周长
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// 方法二 : 遇到相邻的陆地总周长就-2
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int landSum = 0; // 陆地数量
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int cover = 0; // 相邻陆地数量
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for (int i = 0; i < grid.length; i++) {
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for (int j = 0; j < grid[0].length; j++) {
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if (grid[i][j] == 1) {
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landSum++;
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// 统计上面和左边的相邻陆地
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if(i - 1 >= 0 && grid[i-1][j] == 1) cover++;
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if(j - 1 >= 0 && grid[i][j-1] == 1) cover++;
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}
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}
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}
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return landSum * 4 - cover * 2;
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}
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}
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```
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Python：
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### 解法1:
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扫描每个cell,如果当前位置为岛屿 grid[i][j] == 1， 从当前位置判断四边方向，如果边界或者是水域，证明有边界存在，res矩阵的对应cell加一。
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```python
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class Solution:
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def islandPerimeter(self, grid: List[List[int]]) -> int:
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m = len(grid)
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n = len(grid[0])
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# 创建res二维素组记录答案
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res = [[0] * n for j in range(m)]
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for i in range(m):
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for j in range(len(grid[i])):
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# 如果当前位置为水域，不做修改或reset res[i][j] = 0
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if grid[i][j] == 0:
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res[i][j] = 0
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# 如果当前位置为陆地，往四个方向判断，update res[i][j]
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elif grid[i][j] == 1:
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if i == 0 or (i > 0 and grid[i-1][j] == 0):
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res[i][j] += 1
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if j == 0 or (j >0 and grid[i][j-1] == 0):
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res[i][j] += 1
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if i == m-1 or (i < m-1 and grid[i+1][j] == 0):
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res[i][j] += 1
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if j == n-1 or (j < n-1 and grid[i][j+1] == 0):
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res[i][j] += 1
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# 最后求和res矩阵，这里其实不一定需要矩阵记录，可以设置一个variable res 记录边长，舍矩阵无非是更加形象而已
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ans = sum([sum(row) for row in res])
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return ans
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```
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Go：
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```go
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func islandPerimeter(grid [][]int) int {
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m, n := len(grid), len(grid[0])
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res := 0
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for i := 0; i < m; i++ {
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for j := 0; j < n; j++ {
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if grid[i][j] == 1 {
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res += 4
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// 上下左右四个方向
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if i > 0 && grid[i-1][j] == 1 {res--} // 上边有岛屿
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if i < m-1 && grid[i+1][j] == 1 {res--} // 下边有岛屿
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if j > 0 && grid[i][j-1] == 1 {res--} // 左边有岛屿
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if j < n-1 && grid[i][j+1] == 1 {res--} // 右边有岛屿
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}
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}
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}
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return res
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}
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```
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JavaScript：
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```javascript
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//解法一
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var islandPerimeter = function(grid) {
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// 上下左右 4 个方向
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const dirx = [-1, 1, 0, 0], diry = [0, 0, -1, 1];
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const m = grid.length, n = grid[0].length;
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let res = 0; //岛屿周长
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for(let i = 0; i < m; i++){
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for(let j = 0; j < n; j++){
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if(grid[i][j] === 1){
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for(let k = 0; k < 4; k++){ //上下左右四个方向
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// 计算周边坐标的x,y
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let x = i + dirx[k];
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let y = j + diry[k];
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// 四个方向扩展的新位置是水域或者越界就会为周长贡献1
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if(x < 0 // i在边界上
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|| x >= m // i在边界上
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|| y < 0 // j在边界上
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|| y >= n // j在边界上
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|| grid[x][y] === 0){ // (x,y)位置是水域
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res++;
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continue;
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}
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}
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}
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}
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}
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return res;
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};
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//解法二
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var islandPerimeter = function(grid) {
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let sum = 0; // 陆地数量
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let cover = 0; // 相邻数量
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for(let i = 0; i < grid.length; i++){
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for(let j = 0; j <grid[0].length; j++){
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if(grid[i][j] === 1){
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sum++;
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// 统计上边相邻陆地
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if(i - 1 >= 0 && grid[i-1][j] === 1) cover++;
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// 统计左边相邻陆地
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if(j - 1 >= 0 && grid[i][j-1] === 1) cover++;
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// 为什么没统计下边和右边？ 因为避免重复计算
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}
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}
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}
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return sum * 4 - cover * 2;
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};
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```
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<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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