给定一个大小为N * 2的二维数组Knights[][] ,格式为{ X, Y } 的每一行代表骑士的坐标,数组pawn[]代表N * N棋盘中一个棋子的坐标,任务是找出棋盘上攻击棋子的骑士数。
例子:
Input: knights[][] = { { 0, 4 }, { 4, 5 }, { 1, 4 }, { 3, 1 } }, pawn[] = { 2, 3 }
Output: 2
Explanation:
The knights present at coordinate { { 0, 4 }, { 3, 1 } } are attacking the pawn.
Therefore, the required output is 2.
Input: knights[][] = { { 4, 6 }, { 7, 5 }, { 5, 5 } }, pawn[] = { 6, 7 }
Output: 3
Explanation:
The knights present at coordinate { { 4, 6 }, { 7, 5 }, { 5, 5 } } are attacking the pawn.
Therefore, the required output is 3.
方法:按照下面给出的步骤解决问题
- 初始化一个变量,比如cntKnights ,以存储攻击 pawn 的骑士数。
- 使用变量i遍历Knights[][]数组,对于每个数组元素Knights[i] ,检查数组{ (knights[i][0] – pawn[0]), (knights[i][1] – pawn[1]) }是否等于{ 1, 2 }或{ 2, 1 } 。如果发现为真,则将cntKnights的值增加1 。
- 最后,打印cntKnights的值。
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to count the knights that are
// attacking the pawn in an M * M board
int cntKnightsAttackPawn(int knights[][2],
int pawn[], int M)
{
// Stores count of knights that
// are attacking the pawn
int cntKnights = 0;
// Traverse the knights[][] array
for (int i = 0; i < M; i++) {
// Stores absolute difference of X
// co-ordinate of i-th knight and pawn
int X = abs(knights[i][0]
- pawn[0]);
// Stores absolute difference of Y
// co-ordinate of i-th knight and pawn
int Y = abs(knights[i][1]
- pawn[1]);
// If X is 1 and Y is 2 or
// X is 2 and Y is 1
if ((X == 1 && Y == 2)
|| (X == 2 && Y == 1)) {
// Update cntKnights
cntKnights++;
}
}
return cntKnights;
}
// Driver Code
int main()
{
int knights[][2] = { { 0, 4 }, { 4, 5 },
{ 1, 4 }, { 3, 1 } };
int pawn[] = { 2, 3 };
// Stores total count of knights
int M = sizeof(knights)
/ sizeof(knights[0]);
cout << cntKnightsAttackPawn(
knights, pawn, M);
return 0;
}
Java
// Java program to implement
// the above approach
import java.io.*;
import java.lang.Math;
class GFG{
// Function to count the knights that are
// attacking the pawn in an M * M board
static int cntKnightsAttackPawn(int knights[][],
int pawn[], int M)
{
// Stores count of knights that
// are attacking the pawn
int cntKnights = 0;
// Traverse the knights[][] array
for(int i = 0; i < M; i++)
{
// Stores absolute difference of X
// co-ordinate of i-th knight and pawn
int X = Math.abs(knights[i][0] - pawn[0]);
// Stores absolute difference of Y
// co-ordinate of i-th knight and pawn
int Y = Math.abs(knights[i][1] - pawn[1]);
// If X is 1 and Y is 2 or
// X is 2 and Y is 1
if ((X == 1 && Y == 2) ||
(X == 2 && Y == 1))
{
// Update cntKnights
cntKnights++;
}
}
return cntKnights;
}
// Driver code
public static void main(String[] args)
{
int[][] knights = { { 0, 4 }, { 4, 5 },
{ 1, 4 }, { 3, 1 } };
int[] pawn = new int[]{2, 3};
// Stores total count of knights
int M = knights.length;
System.out.println(cntKnightsAttackPawn(
knights, pawn, M));
}
}
// This code is contributed by vandanakillari54935
Python3
# Python program to implement
# the above approach
# Function to count the knights that are
# attacking the pawn in an M * M board
def cntKnightsAttackPawn(knights, pawn, M):
# Stores count of knights that
# are attacking the pawn
cntKnights = 0;
# Traverse the knights array
for i in range(M):
# Stores absolute difference of X
# co-ordinate of i-th knight and pawn
X = abs(knights[i][0] - pawn[0]);
# Stores absolute difference of Y
# co-ordinate of i-th knight and pawn
Y = abs(knights[i][1] - pawn[1]);
# If X is 1 and Y is 2 or
# X is 2 and Y is 1
if ((X == 1 and Y == 2) or (X == 2 and Y == 1)):
# Update cntKnights
cntKnights += 1;
return cntKnights;
# Driver code
if __name__ == '__main__':
knights = [[0, 4], [4, 5], [1, 4], [3, 1]];
pawn = [2, 3];
# Stores total count of knights
M = len(knights);
print(cntKnightsAttackPawn(knights, pawn, M));
# This code is contributed by Amit Katiyar
C#
// C# program to implement
// the above approach
using System;
class GFG
{
// Function to count the knights that are
// attacking the pawn in an M * M board
static int cntKnightsAttackPawn(int[,] knights, int[] pawn, int M)
{
// Stores count of knights that
// are attacking the pawn
int cntKnights = 0;
// Traverse the knights[][] array
for (int i = 0; i < M; i++) {
// Stores absolute difference of X
// co-ordinate of i-th knight and pawn
int X = Math.Abs(knights[i, 0] - pawn[0]);
// Stores absolute difference of Y
// co-ordinate of i-th knight and pawn
int Y = Math.Abs(knights[i, 1] - pawn[1]);
// If X is 1 and Y is 2 or
// X is 2 and Y is 1
if ((X == 1 && Y == 2)
|| (X == 2 && Y == 1)) {
// Update cntKnights
cntKnights++;
}
}
return cntKnights;
}
// Driver code
static void Main()
{
int[,] knights = {{ 0, 4 }, { 4, 5 }, { 1, 4 }, { 3, 1 }};
int[] pawn = {2, 3};
// Stores total count of knights
int M = knights.GetLength(0);
Console.WriteLine(cntKnightsAttackPawn(knights, pawn, M));
}
}
// This code is contributed by divyeshrabadiya07
Javascript
2
时间复杂度: O(M),其中M是骑士总数
辅助空间: O(1)
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