划分正负，不与 0 比较
原文:https://www . geesforgeks . org/partition-negative-positive-不带比较-0/
给定一个 n 个整数的数组，包括负数和正数，将它们分成两个不同的数组，不要将任何元素与 0 进行比较。
示例:
Input : arr[] = [1, -2, 6, -7, 8]
Output : a[] = [1, 6, 8]
b[] = [-2, -7]
算法:
- 初始化两个空向量。将数组的第一个元素推入两个向量中的任何一个。假设第一个向量。让它用 x 表示。
- 对于每隔一个元素，arr[1]到 arr[n-1]，检查它的符号和 x 的符号是否相同。如果符号相同，则将元素推入同一向量。否则，将元素推入另一个向量。
- 完成两个向量的遍历后，打印两个向量。
如何检查它们的标志是否相反？ 让检查的整数用 x 和 y 表示，负数的符号位是 1，正数的符号位是 0。当且仅当 x 和 y 的符号相反时，它们的异或符号位为 1。换句话说，当且仅当 x 和 y 符号相反时，x 和 y 的异或将是负数。
卡片打印处理机（Card Print Processor 的缩写）
// CPP program to rearrange positive and
// negative numbers without comparison
// with 0.
#include <bits/stdc++.h>
using namespace std;
bool oppositeSigns(int x, int y)
{
return ((x ^ y) < 0);
}
void partitionNegPos(int arr[], int n)
{
vector<int> a, b;
// Push first element to a.
a.push_back(arr[0]);
// Now put all elements of same sign
// in a[] and opposite sign in b[]
for (int i = 1; i < n; i++) {
if (oppositeSigns(a[0], arr[i]))
b.push_back(arr[i]);
else
a.push_back(arr[i]);
}
// Print a[] and b[]
for (int i = 0; i < a.size(); i++)
cout << a[i] << ' ';
cout << '\n';
for (int i = 0; i < b.size(); i++)
cout << b[i] << ' ';
}
int main()
{
int arr[] = { 1, -2, 6, -7, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
partitionNegPos(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to rearrange positive and
// negative numbers without comparison
// with 0.
import java.util.*;
class GFG
{
static boolean oppositeSigns(int x, int y)
{
return ((x ^ y) < 0);
}
static void partitionNegPos(int arr[], int n)
{
Vector<Integer> a = new Vector<Integer>();
Vector<Integer> b = new Vector<Integer>();
// Push first element to a.
a.add(arr[0]);
// Now put all elements of same sign
// in a[] and opposite sign in b[]
for (int i = 1; i < n; i++)
{
if (oppositeSigns(a.get(0), arr[i]))
b.add(arr[i]);
else
a.add(arr[i]);
}
// Print a[] and b[]
for (int i = 0; i < a.size(); i++)
System.out.print(a.get(i) + " ");
System.out.println("");
for (int i = 0; i < b.size(); i++)
System.out.print(b.get(i) + " ");
}
public static void main(String[] args)
{
int arr[] = { 1, -2, 6, -7, 8 };
int n = arr.length;
partitionNegPos(arr, n);
}
}
// This code has been contributed by 29AjayKumar
Python 3
# Python3 program to rearrange positive
# and negative numbers without comparison
# with 0.
def oppositeSigns(x, y):
return ((x ^ y) < 0)
def partitionNegPos(arr, n):
a = []
b = []
# Push first element to a.
a = a + [arr[0]]
# Now put all elements of same sign
# in a[] and opposite sign in b[]
for i in range(1, n) :
if (oppositeSigns(a[0], arr[i])):
b = b + [arr[i]]
else:
a = a + [arr[i]]
# Print a[] and b[]
for i in range(0, len(a)):
print(a[i], end = ' ')
print("")
for i in range(0, len(b)):
print(b[i], end = ' ')
# Driver code
arr = [1, -2, 6, -7, 8 ]
n = len(arr)
partitionNegPos(arr, n)
# This code is contributed by Smitha
C
// C# program to rearrange positive and
// negative numbers without comparison
// with 0.
using System;
using System.Collections.Generic;
class GFG
{
static bool oppositeSigns(int x, int y)
{
return ((x ^ y) < 0);
}
static void partitionNegPos(int []arr, int n)
{
List<int> a = new List<int> ();
List<int> b = new List<int> ();
// Push first element to a.
a.Add(arr[0]);
// Now put all elements of same sign
// in a[] and opposite sign in b[]
for (int i = 1; i < n; i++)
{
if (oppositeSigns(a[0], arr[i]))
b.Add(arr[i]);
else
a.Add(arr[i]);
}
// Print a[] and b[]
for (int i = 0; i < a.Count; i++)
Console.Write(a[i] + " ");
Console.WriteLine("");
for (int i = 0; i < b.Count; i++)
Console.Write(b[i] + " ");
}
// Driver code
public static void Main()
{
int []arr = { 1, -2, 6, -7, 8 };
int n = arr.Length;
partitionNegPos(arr, n);
}
}
/* This code contributed by PrinciRaj1992 */
java 描述语言
<script>
// Javascript program to rearrange positive and
// negative numbers without comparison
// with 0.
function oppositeSigns(x, y)
{
return ((x ^ y) < 0);
}
function partitionNegPos(arr, n)
{
var a = [], b = [];
// Push first element to a.
a.push(arr[0]);
// Now put all elements of same sign
// in a[] and opposite sign in b[]
for (var i = 1; i < n; i++) {
if (oppositeSigns(a[0], arr[i]))
b.push(arr[i]);
else
a.push(arr[i]);
}
// Print a[] and b[]
for (var i = 0; i < a.length; i++)
document.write( a[i] + ' ');
document.write( "<br>" );
for (var i = 0; i < b.length; i++)
document.write( b[i] + ' ');
}
var arr = [1, -2, 6, -7, 8];
var n = arr.length;
partitionNegPos(arr, n);
</script>
输出:
1 6 8
-2 -7
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