与矩阵中给定的和配对
给定一个 NxM 矩阵和一个和 s，任务是检查矩阵中是否存在给定和的对。 例 :
Input: mat[N][M] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16}};
sum = 31
Output: YES
Input: mat[N][M] = {{1, 2, 3, 4},
{5, 6, 7, 8}};
sum = 150
Output: NO
进场:
- 取一个散列来存储散列中矩阵的所有元素。
- 开始遍历矩阵，并在遍历时检查散列中是否存在 abs(sum-matrix_element)。
- 如果存在，则返回 true，否则将当前矩阵元素插入散列。
- 如果遍历了矩阵的所有元素，但没有找到对，则返回 false。
以下是上述方法的实现:
C++
// CPP code to check for pair with given sum
#include <bits/stdc++.h>
using namespace std;
#define N 4
#define M 4
// Function to check if a pair with given sum
// exists in the matrix
bool isPairWithSum(int mat[N][M], int sum)
{
// hash to store elements
unordered_set<int> s;
// looping through elements
// if present in the matrix
// return true, else push
// the element in matrix
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
if (s.find(sum - mat[i][j]) != s.end()) {
return true;
}
else {
s.insert(mat[i][j]);
}
}
}
return false;
}
// Driver code
int main()
{
// Input matrix
int mat[N][M] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
// given sum
int sum = 11;
if (isPairWithSum(mat, sum)) {
cout << "YES" << endl;
}
else
cout << "NO" << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to check for pair
// with given sum
import java.util.*;
class GFG
{
// Function to check if a pair with
// given sum exists in the matrix
static final int N = 4;
static final int M = 4;
static boolean isPairWithSum(int [][]mat,
int sum)
{
// hash to store elements
Set<Integer> s = new HashSet<Integer>();
// looping through elements
// if present in the matrix
// return true, else push
// the element in matrix
for (int i = 0; i < N; i++)
{
for (int j = 0; j < M; j++)
{
if (s.contains(sum - mat[i][j]))
{
return true;
}
else
{
s.add(mat[i][j]);
}
}
}
return false;
}
// Driver code
public static void main(String []args)
{
// Input matrix
int [][]mat = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
// given sum
int sum = 11;
if (isPairWithSum(mat, sum))
{
System.out.println("YES");
}
else
System.out.println("NO");
}
}
// This code is contributed by ihritik
C
// C# code to check for pair
// with given sum
using System;
using System.Collections.Generic;
class GFG
{
// Function to check if a pair with
// given sum exists in the matrix
static readonly int N = 4;
static readonly int M = 4;
static bool isPairWithSum(int [,]mat,
int sum)
{
// hash to store elements
HashSet<int> s = new HashSet<int>();
// looping through elements
// if present in the matrix
// return true, else push
// the element in matrix
for (int i = 0; i < N; i++)
{
for (int j = 0; j < M; j++)
{
if (s.Contains(sum - mat[i, j]))
{
return true;
}
else
{
s.Add(mat[i, j]);
}
}
}
return false;
}
// Driver code
public static void Main(String []args)
{
// Input matrix
int [,]mat = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
// given sum
int sum = 11;
if (isPairWithSum(mat, sum))
{
Console.WriteLine("YES");
}
else
Console.WriteLine("NO");
}
}
// This code contributed by Rajput-Ji
Python 3
# python code to check for pair with given sum
N= 4
M= 4
# Function to check if a pair with given sum
# exists in the matrix
def isPairWithSum(mat, sum):
# hash to store elements
s = set()
# looping through elements
# if present in the matrix
# return true, else push
# the element in matrix
for i in range(N):
for j in range(M):
if (sum - mat[i][j]) in s :
return True
else :
s.add(mat[i][j])
return False
# Driver code
if __name__ == '__main__':
# Input matrix
mat = [[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8] ,
[ 9, 10, 11, 12] ,
[13, 14, 15, 16]]
# given sum
sum = 11
if (isPairWithSum(mat, sum)) :
print("YES")
else:
print("NO")
# This code is contributed by AbhiThakur
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP code to check for pair with given sum
$N = 4;
$M = 4;
// Function to check if a pair with given sum
// exists in the matrix
function isPairWithSum(&$mat, $sum)
{
global $N,$M;
// array to store elements
$s = array();
// looping through elements
// if present in the matrix
// return true, else push
// the element in matrix
for ($i = 0; $i < $N; $i++) {
for ($j = 0; $j < $M; $j++) {
if (in_array($sum - $mat[$i][$j],$s) != end($s)) {
return true;
}
else {
array_push($s,$mat[$i][$j]);
}
}
}
return false;
}
// Driver code
// Input matrix
$mat = array(array( 1, 2, 3, 4 ),
array( 5, 6, 7, 8 ),
array( 9, 10, 11, 12 ),
array(13, 14, 15, 16 ));
// given sum
$sum = 11;
if (isPairWithSum($mat, $sum)) {
echo "YES" ."\n";
}
else
echo "NO" ."\n";
return 0;
?>
java 描述语言
<script>
// Javascript code to check for pair
// with given sum
// Function to check if a pair with
// given sum exists in the matrix
let N = 4;
let M = 4;
function isPairWithSum(mat,sum)
{
// hash to store elements
let s = new Set();
// looping through elements
// if present in the matrix
// return true, else push
// the element in matrix
for (let i = 0; i < N; i++)
{
for (let j = 0; j < M; j++)
{
if (s.has(sum - mat[i][j]))
{
return true;
}
else
{
s.add(mat[i][j]);
}
}
}
return false;
}
// Driver code
// Input matrix
let mat = [[ 1, 2, 3, 4 ],
[ 5, 6, 7, 8] ,
[ 9, 10, 11, 12] ,
[13, 14, 15, 16]] ;
// given sum
let sum = 11;
if (isPairWithSum(mat, sum))
{
document.write("YES");
}
else
document.write("NO");
// This code is contributed by rag2127
</script>
Output:
YES
时间复杂度 : O(N*M)
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