查找位置元素的数量
给定一个整数矩阵，任务是找出位置元素的个数。位置元素是行或列中最小或最大的元素。
示例:
输入:a = {{1，3，4}，{5，2，9}，{8，7，6}} 输出:7 共有 7 个元素最小元素为 1，2，6 和 4。最大元素是 9，8 和 7。
输入:a = {{1，1}，{1，1}，{1，1}} 输出:6
来源 : 高盛面试集 思路是存储每一行每一列的最大值和最小值，然后检查所需条件。
下面是上述方法的实现。
C++
// CPP program to find positional elements in
// a matrix.
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
int countPositional(int a[][MAX], int m, int n)
{
// rwomax[i] is going to store maximum of
// i-th row and other arrays have similar
// meaning
int rowmax[m], rowmin[m];
int colmax[n], colmin[n];
// Find rminn and rmaxx for every row
for (int i = 0; i < m; i++) {
int rminn = INT_MAX;
int rmaxx = INT_MIN;
for (int j = 0; j < n; j++) {
if (a[i][j] > rmaxx)
rmaxx = a[i][j];
if (a[i][j] < rminn)
rminn = a[i][j];
}
rowmax[i] = rmaxx;
rowmin[i] = rminn;
}
// Find cminn and cmaxx for every column
for (int j = 0; j < n; j++) {
int cminn = INT_MAX;
int cmaxx = INT_MIN;
for (int i = 0; i < m; i++) {
if (a[i][j] > cmaxx)
cmaxx = a[i][j];
if (a[i][j] < cminn)
cminn = a[i][j];
}
colmax[j] = cmaxx;
colmin[j] = cminn;
}
// Check for optimal element
int count = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if ((a[i][j] == rowmax[i])
|| (a[i][j] == rowmin[i])
|| (a[i][j] == colmax[j])
|| (a[i][j] == colmin[j])) {
count++;
}
}
}
return count;
}
// Driver code
int main()
{
int a[][MAX] = { { 1, 3, 4 },
{ 5, 2, 9 },
{ 8, 7, 6 } };
int m = 3, n = 3;
cout << countPositional(a, m, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find positional elements in
// a matrix.
class GfG {
static int MAX = 100;
static int countPositional(int a[][], int m, int n)
{
// rwomax[i] is going to store maximum of
// i-th row and other arrays have similar
// meaning
int rowmax[] = new int[m];
int rowmin[] = new int[m];
int colmax[] = new int[n];
int colmin[] = new int[n];
// Find rminn and rmaxx for every row
for (int i = 0; i < m; i++) {
int rminn = Integer.MAX_VALUE;
int rmaxx = Integer.MIN_VALUE;
for (int j = 0; j < n; j++) {
if (a[i][j] > rmaxx)
rmaxx = a[i][j];
if (a[i][j] < rminn)
rminn = a[i][j];
}
rowmax[i] = rmaxx;
rowmin[i] = rminn;
}
// Find cminn and cmaxx for every column
for (int j = 0; j < n; j++) {
int cminn = Integer.MAX_VALUE;
int cmaxx = Integer.MIN_VALUE;
for (int i = 0; i < m; i++) {
if (a[i][j] > cmaxx)
cmaxx = a[i][j];
if (a[i][j] < cminn)
cminn = a[i][j];
}
colmax[j] = cmaxx;
colmin[j] = cminn;
}
// Check for optimal element
int count = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if ((a[i][j] == rowmax[i])
|| (a[i][j] == rowmin[i])
|| (a[i][j] == colmax[j])
|| (a[i][j] == colmin[j])) {
count++;
}
}
}
return count;
}
public static void main(String[] args)
{
int a[][] = new int[][] { { 1, 3, 4 }, { 5, 2, 9 }, { 8, 7, 6 } };
int m = 3, n = 3;
System.out.println(countPositional(a, m, n));
}
}
Python 3
# Python3 program to find positional elements in a matrix.
import sys
MAX = 100
def countPositional(a, m, n):
# rwomax[i] is going to store maximum of
# i-th row and other arrays have similar
# meaning
rowmax = [0] * m
rowmin = [0] * m
colmax = [0] * n
colmin = [0] * n
# Find rminn and rmaxx for every row
for i in range(m) :
rminn = sys.maxsize
rmaxx = -sys.maxsize
for j in range(n) :
if (a[i][j] > rmaxx) :
rmaxx = a[i][j]
if (a[i][j] < rminn) :
rminn = a[i][j]
rowmax[i] = rmaxx
rowmin[i] = rminn
# Find cminn and cmaxx for every column
for j in range(n) :
cminn = sys.maxsize
cmaxx = -sys.maxsize
for i in range(m) :
if (a[i][j] > cmaxx) :
cmaxx = a[i][j]
if (a[i][j] < cminn) :
cminn = a[i][j]
colmax[j] = cmaxx
colmin[j] = cminn
# Check for optimal element
count = 0
for i in range(m) :
for j in range(n) :
if ((a[i][j] == rowmax[i]) or (a[i][j] == rowmin[i])
or (a[i][j] == colmax[j])
or (a[i][j] == colmin[j])) :
count += 1
return count
# Driver code
a = [ [ 1, 3, 4 ], [ 5, 2, 9 ], [ 8, 7, 6 ] ]
m, n = 3, 3
print(countPositional(a, m, n))
# This code is contributed by divyeshrabadiya07
C
// C# program to find positional elements in
using System;
class GFG {
static int countPositional(int[, ] a, int m, int n)
{
// rwomax[i] is going to store maximum of
// i-th row and other arrays have similar
// meaning
int[] rowmax = new int[m];
int[] rowmin = new int[m];
int[] colmax = new int[n];
int[] colmin = new int[n];
// Find rminn and rmaxx for every row
for (int i = 0; i < m; i++) {
int rminn = int.MaxValue;
int rmaxx = int.MinValue;
for (int j = 0; j < n; j++) {
if (a[i, j] > rmaxx)
rmaxx = a[i, j];
if (a[i, j] < rminn)
rminn = a[i, j];
}
rowmax[i] = rmaxx;
rowmin[i] = rminn;
}
// Find cminn and cmaxx for every column
for (int j = 0; j < n; j++) {
int cminn = int.MaxValue;
int cmaxx = int.MinValue;
for (int i = 0; i < m; i++) {
if (a[i, j] > cmaxx)
cmaxx = a[i, j];
if (a[i, j] < cminn)
cminn = a[i, j];
}
colmax[j] = cmaxx;
colmin[j] = cminn;
}
// Check for optimal element
int count = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if ((a[i, j] == rowmax[i])
|| (a[i, j] == rowmin[i])
|| (a[i, j] == colmax[j])
|| (a[i, j] == colmin[j])) {
count++;
}
}
}
return count;
}
// Driver Code
static public void Main()
{
int[, ] a = new int[, ] { { 1, 3, 4 }, { 5, 2, 9 }, { 8, 7, 6 } };
int m = 3, n = 3;
Console.WriteLine(countPositional(a, m, n));
}
}
// This code is contributed by Tushil.
java 描述语言
<script>
// Javascript program to find positional
// elements in a matrix.
let MAX = 100;
function countPositional(a, m, n)
{
// rwomax[i] is going to store maximum of
// i-th row and other arrays have similar
// meaning
let rowmax = new Array(m);
let rowmin = new Array(m);
let colmax = new Array(n);
let colmin = new Array(n);
// Find rminn and rmaxx for every row
for(let i = 0; i < m; i++)
{
let rminn = Number.MAX_VALUE;
let rmaxx = Number.MIN_VALUE;
for(let j = 0; j < n; j++)
{
if (a[i][j] > rmaxx)
rmaxx = a[i][j];
if (a[i][j] < rminn)
rminn = a[i][j];
}
rowmax[i] = rmaxx;
rowmin[i] = rminn;
}
// Find cminn and cmaxx for every column
for(let j = 0; j < n; j++)
{
let cminn = Number.MAX_VALUE;
let cmaxx = Number.MIN_VALUE;
for(let i = 0; i < m; i++)
{
if (a[i][j] > cmaxx)
cmaxx = a[i][j];
if (a[i][j] < cminn)
cminn = a[i][j];
}
colmax[j] = cmaxx;
colmin[j] = cminn;
}
// Check for optimal element
let count = 0;
for(let i = 0; i < m; i++)
{
for(let j = 0; j < n; j++)
{
if ((a[i][j] == rowmax[i]) ||
(a[i][j] == rowmin[i]) ||
(a[i][j] == colmax[j]) ||
(a[i][j] == colmin[j]))
{
count++;
}
}
}
return count;
}
// Driver code
let a = [ [ 1, 3, 4 ],
[ 5, 2, 9 ],
[ 8, 7, 6 ] ];
let m = 3, n = 3;
document.write(countPositional(a, m, n));
// This code is contributed by suresh07
</script>
Output:
7
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