使用给定的矩形块可以形成的最大正方形
给定一个正整数的数组 arr[] ，其中数组的每个元素代表矩形块的长度。任务是找到可以用矩形块形成的正方形的最大长度。
示例:
输入: arr[] = {3，2，1，5，2，4} 输出: 3 说明: 利用长度为 3，5，4 的矩形块，边长为 3 的正方形可构造如下:
输入: arr[] = {1，2，3 } T3】输出: 2
进场:
- 按降序对给定数组进行排序。
- 将最大边长初始化为 0。
- 遍历数组 arr[] ，如果 arr[i] >最大长度，则增加最大长度，并为下一次迭代检查该条件。
- 如果上述条件不满足，则中断循环并打印最大长度。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find maximum side
// length of square
void maxSide(int a[], int n)
{
int sideLength = 0;
// Sort array in asc order
sort(a, a + n);
// Traverse array in desc order
for (int i = n - 1; i >= 0; i--) {
if (a[i] > sideLength) {
sideLength++;
}
else {
break;
}
}
cout << sideLength << endl;
}
// Driver Code
int main()
{
int N = 6;
// Given array arr[]
int arr[] = { 3, 2, 1, 5, 2, 4 };
// Function Call
maxSide(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.Arrays;
class GFG{
// Function to find maximum side
// length of square
static void maxSide(int a[], int n)
{
int sideLength = 0;
// Sort array in asc order
Arrays.sort(a);
// Traverse array in desc order
for(int i = n - 1; i >= 0; i--)
{
if (a[i] > sideLength)
{
sideLength++;
}
else
{
break;
}
}
System.out.println(sideLength);
}
// Driver code
public static void main (String[] args)
{
int N = 6;
// Given array arr[]
int arr[] = new int[]{ 3, 2, 1,
5, 2, 4 };
// Function Call
maxSide(arr, N);
}
}
// This code is contributed by Pratima Pandey
Python 3
# Python3 program for the above approach
# Function to find maximum side
# length of square
def maxSide(a, n):
sideLength = 0
# Sort array in asc order
a.sort
# Traverse array in desc order
for i in range(n - 1, -1, -1):
if (a[i] > sideLength):
sideLength += 1
else:
break
print(sideLength)
# Driver code
N = 6
# Given array arr[]
arr = [ 3, 2, 1, 5, 2, 4 ]
# Function Call
maxSide(arr, N)
# This code is contributed by divyeshrabadiya07
C
// C# program for the above approach
using System;
class GFG{
// Function to find maximum side
// length of square
static void maxSide(int []a, int n)
{
int sideLength = 0;
// Sort array in asc order
Array.Sort(a);
// Traverse array in desc order
for(int i = n - 1; i >= 0; i--)
{
if (a[i] > sideLength)
{
sideLength++;
}
else
{
break;
}
}
Console.Write(sideLength);
}
// Driver code
public static void Main()
{
int N = 6;
// Given array arr[]
int []arr = new int[]{ 3, 2, 1,
5, 2, 4 };
// Function Call
maxSide(arr, N);
}
}
// This code is contributed by Code_Mech
java 描述语言
<script>
// Javascript program for the above approach
// Function to find maximum side
// length of square
function maxSide( a, n) {
let sideLength = 0;
// Sort array in asc order
a.sort();
// Traverse array in desc order
for ( i = n - 1; i >= 0; i--) {
if (a[i] > sideLength) {
sideLength++;
} else {
break;
}
}
document.write(sideLength);
}
// Driver code
let N = 6;
// Given array arr
let arr = [3, 2, 1, 5, 2, 4 ];
// Function Call
maxSide(arr, N);
// This code contributed by aashish1995
</script>
Output:
3
时间复杂度:O(N * log N) T5】辅助空间: O(1)
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