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C++或Go求矩陣?yán)锏膷u嶼的數(shù)量詳解

更新時間：2021年09月09日 14:30:35 作者：sanqima

這篇文章主要介紹了C++和go實現(xiàn)LeetCode(200.島嶼的數(shù)量),本篇文章通過簡要的案例,講解了該項技術(shù)的了解與使用,以下就是詳細(xì)內(nèi)容,需要的朋友可以參考下

1、C++實現(xiàn)

//island.cpp

#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;
//判斷坐標(biāo)(r,c)是否存在網(wǎng)絡(luò)中
bool inArea(vector<vector<char>>& grid, int r, int c) {
	bool bRow = (r >= 0) && (r < (int)grid.size());
	bool bCol = (c >= 0) && (c < (int)grid[0].size());
	return bRow && bCol;
}
//void dfs(int[][] grid, int r, int c) {
void dfs(vector<vector<char>>& grid, int r,int c){
	//判斷base case
	//如果坐標(biāo)(r,c)超出了網(wǎng)格范圍，則直接返回
	if (!inArea(grid,r,c)) {
		return;
	}
	//如果不是島嶼，則直接返回
	if (grid[r][c] != '1') {
		return;
	}
	//將原來的"1"改成"0"
	grid[r][c] = '2';
	//訪問上、下、左、右四個相鄰結(jié)點
	dfs(grid, r - 1, c);
	dfs(grid, r + 1, c);
	dfs(grid, r , c-1);
	dfs(grid, r , c+1);
}
//求島嶼的個數(shù)
//時間復(fù)雜度：O(MN)O(MN)，其中 MM 和 NN 分別為行數(shù)和列數(shù)。
//空間復(fù)雜度：O(MN)O(MN)，在最壞情況下，整個網(wǎng)格均為陸地，深度優(yōu)先搜索的深度達到MN。
//
int numIslands(vector<vector<char>>& grid){
	int r = grid.size();
	if (!r)
		return 0;
	int c = grid[0].size();
	int num = 0;
	for (int i = 0; i < r; i++) {
		for (int j = 0; j < c; j++) {
			if (grid[i][j] == '1') {
				++num;
				dfs(grid, i, j);
			}
		}
	}
	return num;
}
int main(){
	//島嶼
	// 1  1  1
	// 0  1  0
	// 1  0  0
	// 1  0  1
	vector<char> row1;
	row1.push_back('1');
	row1.push_back('1');
	row1.push_back('1');
	vector<char> row2;
	row2.push_back('0');
	row2.push_back('1');
	row2.push_back('0');
	vector<char> row3;
	row3.push_back('1');
	row3.push_back('0');
	row3.push_back('0');
	vector<char> row4;
	row4.push_back('1');
	row4.push_back('0');
	row4.push_back('1');
	vector<vector<char>> grid;
	grid.push_back(row1);
	grid.push_back(row2);
	grid.push_back(row3);
	grid.push_back(row4);
	int numLands = numIslands(grid);
	cout << "numLands= " << numLands << endl;
	system("pause");
	return 0;
}

效果如下：

圖(1) 孤島的個數(shù)

2、go語言實現(xiàn)

//island.go

package main
import "fmt"
func numIslands(grid [][]byte) int {
	nums := 0
	for i:=0; i<len(grid); i++ {
		for j:=0; j<len(grid[0]); j++ {
			if grid[i][j] == '1' {
				DFS(&grid,i,j)
				nums++
			}
		}
	}
	return nums
}
func DFS(grid *[][]byte, i int, j int) {
	var (
		row = len(*grid)
		col = len((*grid)[0])
	)
	if i<0 || i>=row || j<0 || j>= col {
		return
	}
	if (*grid)[i][j] == '1' {
		(*grid)[i][j] = '2'
		DFS(grid,i-1,j)
		DFS(grid,i+1,j)
		DFS(grid,i,j-1)
		DFS(grid,i,j+1)
	}
}
func main() {
	var grid = make([][]byte, 4)
	grid[0] = []byte{'1','1','1'}
	grid[1] = []byte{'0','1','0'}
	grid[2] = []byte{'1','0','0'}
	grid[3] = []byte{'1','0','1'}
	res := numIslands(grid)
	fmt.Println("numlands=",res)
}

效果如下：