概述

给出两个整数m和n，代表一个m*n的网格。除此以外，还给出了两个二维整数的数组

• guards，其中guards[i] = [rowi , columni]。它表示第i个守卫的位置

• 墙，其中guards[j] = [rowi , columni]。它代表第i个墙的位置

一个警卫可以看到所有的方向

• 北边

• 南边

• 东

• 西

除非有墙的阻挡。所有守卫能看到的牢房都算作有守卫。我们的目标是找到无人看守的牢房数量。

程序

以下是相同的程序

package main

import "fmt"

func countUnguarded(m int, n int, guards [][]int, walls [][]int) int {

	occupancy := make([][]int, m)

	for i := 0; i < m; i++ {
		occupancy[i] = make([]int, n)
	}

	for _, val := range guards {
		i := val[0]
		j := val[1]
		occupancy[i][j] = 2
	}

	for _, val := range walls {
		i := val[0]
		j := val[1]
		occupancy[i][j] = -1
	}

	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if occupancy[i][j] == 2 {
				countUtil(i, j, m, n, &occupancy)
			}

		}
	}

	count := 0
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if occupancy[i][j] == 0 {
				count += 1
			}

		}
	}

	return count
}

func countUtil(x, y, m, n int, occupancy *([][]int)) {

	for i := x + 1; i < m; i++ {
		if (*occupancy)[i][y] == 0 {
			(*occupancy)[i][y] = 1
		}

		if (*occupancy)[i][y] == -1 || (*occupancy)[i][y] == 2 {
			break
		}
	}

	for i := x - 1; i >= 0; i-- {
		if (*occupancy)[i][y] == 0 {
			(*occupancy)[i][y] = 1
		}

		if (*occupancy)[i][y] == -1 || (*occupancy)[i][y] == 2 {
			break
		}
	}

	for i := y + 1; i < n; i++ {
		if (*occupancy)[x][i] == 0 {
			(*occupancy)[x][i] = 1
		}

		if (*occupancy)[x][i] == -1 || (*occupancy)[x][i] == 2 {
			break
		}
	}

	for i := y - 1; i >= 0; i-- {

		if (*occupancy)[x][i] == 0 {
			(*occupancy)[x][i] = 1
		}

		if (*occupancy)[x][i] == -1 || (*occupancy)[x][i] == 2 {
			break
		}
	}

}

func main() {
	output := countUnguarded(4, 6, [][]int{{0, 0}, {1, 1}, {2, 3}}, [][]int{{0, 1}, {2, 2}, {1, 4}})
	fmt.Println(output)

	output = countUnguarded(3, 3, [][]int{{1, 1}}, [][]int{{0, 1}, {1, 0}, {2, 1}, {1, 2}})
	fmt.Println(output)
}

输出

7
4

注意： 请查看我们的Golang高级教程。这个系列的教程是精心设计的，我们试图用例子涵盖所有的概念。本教程是为那些希望获得专业知识和扎实了解Golang的人准备的 -Golang高级教程