136.Single Number

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Given an array of integers,every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement itwithout using extra memory?

第二种不使用额外空间的方法，比第一种慢一点点~

//main.cpp
#pragma once
#include<iostream>
using namespace std;


int singleNumber2(int A[], int n)
{
	int result = *A;
	for (int i = 1; i < n; i++)
		result = result^A[i];
	return result;
}

int singleNumber1(int A[], int n)
{
	for (int i = 1; i < n; i++)
		*A = *A^A[i];
	return *A;
}

void main()
{
	int a[] = { 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 };
	int n = sizeof(a) / sizeof(int);
	cout << singleNumber(a, 11) << endl;
	system("pause");
}

136.Single Number

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原文地址：http://blog.csdn.net/hgqqtql/article/details/43278665