剑指offer中找出出现次数超过一半的数字

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#include<iostream>
#include<cstring>
using namespace std;
bool check(int *numbers,int length,int number)
{
	int times=0;
	for(int i=0;i<length;++i)
	{
		if(numbers[i]==number)
			times++;
	}
	bool ismorethanhalf=true;
	if(times*2<=length)
	{
		ismorethanhalf=false;
	}
	return ismorethanhalf;
}
int morethanhalf(int *numbers,int length)
{
	int result=numbers[0];
	int times=1;
	for(int i=1;i<length;++i)
	{
		/*if(times==0)
		{
			result=numbers[i];
			times=1;
		}*/
		 if(numbers[i]==result)
			times++;
		else
		{times--;
		 if(times==0)
		 {
			 result=numbers[i];
			times=1;
		 }
		 }
	}
	if(!check(numbers,length,result))
		result=0;
	return result; 
}
int main()
{
	//int numbers[6]={2,2,2,3,3,3};
	//cout<<morethanhalf(numbers,6);
	char a[5];
	int b=10;
	itoa(b,a,10);
	cout<<a<<endl;
	cout<<strlen(a)<<endl;
	system("pause");
	return 0;
}

剑指offer中找出出现次数超过一半的数字

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原文地址：http://blog.csdn.net/qq_22335577/article/details/44828857