数组中出现次数超过一半的数字

数组中有一个数字出现了超过数组长度的一般，请找出这个数字。

思路：如果这个数子长度超过一半，那肯定排序后他会出现在middle的位置上！

int Partition(int* num, unsigned int length, int start, int end){
	int pivot = num[end];
	int i = start-1;
	int index;
	for (index = start; index<end; index++){
		if (num[index] < pivot){
			i++;
			swap(num[index],num[i]);
		}
	}
	swap (num[index],num[i++]);
	return i;
}
int moreHalf(int* num, unsigned int length){
	if (num == NULL || length == 0){
		return 0;
		isValid = false;
	}

	if (length == 1)
		return *num;

	int middle = length >> 1;
	int start = 0;
	int end = length-1;
	int index = Partition(num, length, start, end);

	while (index != middle){
		if (index>middle){
			end = index-1;
			index = Partition(num, length, start, end);
		}
		else{
			start = index+1;
			index = Partition(num, length, start, end);
		}
	}

	return num[middle];
}

1. 很明显，这个程序有反例：如果输入数组中出现频率最高的数字都没有达到一半。所以在return之前需要检查是否达到标准

bool checkMoreHalf (int* num, int middle, int length){
	int pivot = num[middle];
	int len = 1;
	for (int index = middle-1; index >= 0; index--){
		if (num[index] == piovt)
			len++;
		else
			break;
	}
	for (int index = middle+1; index<length; index++){
		if (num[index] == piovt)
			len++;
		else
			break;
	}

	if (len * 2 <= length)
		return false;
	return true;
}

2.  计算复杂度O(n)

3.  快速排序的思想很好，常用来求：得到数组中任意第K大的数字