微软面试题目(二)移动数组元素

给定一个整数数组,将数组中小于零的数都放在最左边,等于0的放在中间,小于零的放在最右边。

代码如下:

void swap(int* a,int* b)
{
	*a = *a ^ *b;	//a、b中不同位
	*b = *a ^ *b;	//b = a
	*a = *a ^ *b;	//a = b
}
void ArrangArray(int* StartPos,int* EndPos)
{
	//Step1先将小于零的放在最左边,大于等于0的数不区分,都放在右边
	int* low = StartPos;
	int* high = EndPos;
	while(low < high)
	{
		while( *low < 0 && low < high ) low++;
		while( *high >= 0 && low < high ) high--;
		if(low < high)
			swap(low,high);
	}
	//循环结束时,low一定等于high,且指向大于等于0的数
	//Step2,从low开始,将等于0的数移动到左边,大于0的数移动到右边
	high = EndPos;
	while(low < high)
	{
		while( *low == 0 && low < high ) low++;
		while( *high > 0 && low < high ) high--;
		if(low < high)
			swap(low,high);
	}
}
int main()
{
	int array[10] = {-1,3,0,2,-5,0,-3,4,0,-8};
	ArrangArray(array,array + 9);
}
posted @ 2012-05-13 18:14  DylanTsou  阅读(577)  评论(0编辑  收藏  举报