给定两个有符号整数，编写一个函数，如果给定整数的符号不同，则返回true，否则返回false。例如，函数应该返回true-1和+100，对于-100和-200应该返回false。该函数不应使用任何算术运算符。 设给定的整数为x和y。负数的符号位为1，正数的符号位为0。当x和y的符号相反时，它们的异或将符号位设为1。换句话说，当x和y有相反的符号时，x和y的XOR将是负数。下面的代码使用这种逻辑。
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C++
// C++ Program to Detect // if two integers have opposite signs. #include<iostream> using namespace std; bool oppositeSigns( int x, int y) { return ((x ^ y) < 0); } int main() { int x = 100, y = -100; if (oppositeSigns(x, y) == true ) cout << "Signs are opposite" ; else cout << "Signs are not opposite" ; return 0; } // this code is contributed by shivanisinghss2110 |
C
// C++ Program to Detect // if two integers have opposite signs. #include<stdbool.h> #include<stdio.h> bool oppositeSigns( int x, int y) { return ((x ^ y) < 0); } int main() { int x = 100, y = -100; if (oppositeSigns(x, y) == true ) printf ( "Signs are opposite" ); else printf ( "Signs are not opposite" ); return 0; } |
JAVA
// Java Program to Detect // if two integers have opposite signs. class GFG { static boolean oppositeSigns( int x, int y) { return ((x ^ y) < 0 ); } public static void main(String[] args) { int x = 100 , y = - 100 ; if (oppositeSigns(x, y) == true ) System.out.println( "Signs are opposite" ); else System.out.println( "Signs are not opposite" ); } } // This code is contributed by prerna saini. |
Python3
# Python3 Program to Detect # if two integers have # opposite signs. def oppositeSigns(x, y): return ((x ^ y) < 0 ); x = 100 y = 1 if (oppositeSigns(x, y) = = True ): print ( "Signs are opposite" ) else : print ( "Signs are not opposite" ) # This article is contributed by Prerna Saini. |
C#
// C# Program to Detect // if two integers have // opposite signs. using System; class GFG { // Function to detect signs static bool oppositeSigns( int x, int y) { return ((x ^ y) < 0); } // Driver Code public static void Main() { int x = 100, y = -100; if (oppositeSigns(x, y) == true ) Console.Write( "Signs are opposite" ); else Console.Write( "Signs are not opposite" ); } } // This code is contributed by Nitin Mittal. |
PHP
<?php // PHP Program to Detect if two // integers have opposite signs. function oppositeSigns( $x , $y ) { return (( $x ^ $y ) < 0); } // Driver Code $x = 100; $y = -100; if (oppositeSigns( $x , $y ) == true) echo ( "Signs are opposite" ); else echo ( "Signs are not opposite" ); // This code is contributed by vt_m. ?> |
Javascript
<script> // Javascript Program to Detect // if two integers have opposite signs. function oppositeSigns(x, y) { return ((x ^ y) < 0); } // Driver Code let x = 100, y = -100; if (oppositeSigns(x, y) == true ) document.write( "Signs are opposite" ); else document.write( "Signs are not opposite" ); </script> |
输出：
Signs are opposite
时间复杂性： O（1）
辅助空间： O（1）
资料来源： 检测两个整数是否有相反的符号 我们也可以通过使用两个比较运算符来解决这个问题。请参阅下面的代码。
CPP
bool oppositeSigns( int x, int y) { return (x < 0)? (y >= 0): (y < 0); } |
JAVA
class GFG{ static boolean oppositeSigns( int x, int y) { return (x < 0 )? (y >= 0 ): (y < 0 ); } } // This code contributed by umadevi9616 |
Python3
def oppositeSigns(x, y): return (y > = 0 ) if (x < 0 ) else (y < 0 ); # This code is contributed by shivanisingjss2110 |
C#
using System; class GFG{ static boo oppositeSigns( int x, int y) { return (x < 0)? (y >= 0): (y < 0); } } // This code contributed by shivanisinghss2110 |
Javascript
<script> function oppositeSigns(x, y) { return (x < 0)? (y >= 0): (y < 0); } // This code contributed by shivanisinghss2110 </script> |
时间复杂性： O（1）
第一种方法更有效。第一种方法使用位异或和比较运算符。第二种方法使用两个比较运算符，与比较运算相比，按位异或运算效率更高。 我们也可以使用以下方法。它不使用任何比较运算符。该方法由洪亮提出，gaurav改进。
CPP
bool oppositeSigns( int x, int y) { return ((x ^ y) >> 31); } |
JAVA
import java.io.*; class GFG { static boolean oppositeSigns( int x, int y) { return ((x ^ y) >> 31 ); } } // This code is contributed by shivanisinghss2110 |
Python3
def oppositeSigns(x, y): return ((x ^ y) >> 31 ) # this code is contributed by shivanisinghss2110 |
C#
using System; class GFG { static bool oppositeSigns( int x, int y) { return ((x ^ y) >> 31); } } // This code is contributed by shivanisinghss2110 |
Javascript
<script> function oppositeSigns(x, y) { return ((x ^ y) >> 31); } // This code is contributed by shivanisinghss2110 </script> |
时间复杂性： O（1）
该函数仅适用于整数大小为32位的编译器。表达式基本上使用位运算符“>>”检查（x^y）的符号。如上所述，负数的符号位始终为1。符号位是二进制表示法中最左边的位。所以我们需要检查x^y的第32位（或最左边的位）是否为1。我们将x^y的值右移31，使符号位成为最低有效位。如果符号位为1，（x^y）>>31的值将为1，否则为0。
C++
// C++ Program to detect // if two integers have opposite signs. #include<iostream> using namespace std; bool oppositeSigns( int x, int y) { long long product = 1ll*x*y; return (product<0); } int main() { int x = 100, y = -100; if (oppositeSigns(x, y) == true ) cout << "Signs are opposite" ; else cout << "Signs are not opposite" ; return 0; } // this code is contributed by shinjanpatra |
Javascript
// JavaScript Program to detect // if two integers have opposite signs. function oppositeSigns(x,y) { const product = Number(x)*Number(y); return (product<0); } // driver code let x = 100, y = -100; if (oppositeSigns(x, y) == true ) { console.log( "Signs are opposite" ); } else console.log( "Signs are not opposite" ); // this code is contributed by shinjanpatra |
Python3
# Python Program to detect # if two integers have opposite signs. def oppositeSigns(x,y): product = x * y return (product< 0 ) # driver code x = 100 y = - 100 if (oppositeSigns(x, y) = = True ): print ( "Signs are opposite" ) else : print ( "Signs are not opposite" ) # this code is contributed by shinjanpatra |
方法： 基本的方法是计算两个整数的乘积，我们知道，两个符号相反的整数总是会产生一个负整数，我们只需要知道乘积是否为负。
时间复杂性 ：O（1）
空间复杂性 ：O（1）
如果您发现上述任何代码/算法不正确，请写下评论，或者寻找其他方法来解决相同的问题。
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