Rearrange an array of positive and negative integers

数组重排算法

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本文介绍了一种特殊的数组重排算法，该算法可以在保持原有顺序的前提下，将正数和负数分别集中到数组的一端。通过使用递归和块交换技术，能够有效地降低时间复杂度从O(n^2)至O(nlogn)。

Given an array of positive and negative integers, re-arrange it
so that you have postives on one end and negatives on the other,
BUT retain the original order of appearance.

For eg. 1, 7, -5, 9, -12, 15 => -5, -12, 1, 7, 9, 15

可以使用块交换技术实现题目要求，

1,7，-5 交换为 -5,1,7

然后

-5,1,7,9,-12 交换为 -5,-12,1,7,9

目测时间复杂度是o(n^2)

但是可以用递归将时间复杂度降维nlogn

http://haixiaoyang.wordpress.com/?s=Rearrange

有空可以研究下

//Given an array of positive and negative integers, 
//re-arrange it so that you have postives on one end 
//and negatives on the other, BUT retain the original order of appearance. do it in-place
//e.g. 1, 7, -5, 9, -12, 15 => -5, -12, 1, 7, 9, 15

//When considering optimization from O(n^2) to O(nlogn),use divided & conquer
void swapRange(int a[], int nBeg, int nEnd)
{
	assert(a && nBeg <= nEnd);

	while (nBeg < nEnd)
		swap(a[nBeg++], a[nEnd--]);
}

//solution:
//e.g. in any stage: ---- "+++ --" ++++, reverse middle part
// ==> ---- "--" "+++" ++++, reverse middle 2 parts seperatly to keep "stable"
void ReArrange(int a[], int n)
{
	assert(a && n > 0);

	if (n <= 1) return;

	ReArrange(a, n/2);
	ReArrange(a + n/2, n - n/2); //pitfall, notice both parameters

	int nLft = 0;
	while (a[nLft] < 0) 
		nLft++;
	int nRgt = n-1;
	while (a[nRgt] >= 0)
		nRgt--;
	
	//Very important, no need to swap under this situation, return
	if (nRgt <= nLft) return; 

	swapRange(a, nLft, nRgt);

	int nBegRgt = nLft;
	while (a[nBegRgt] < 0) //no need to use "&& nBegRgt < n"
		nBegRgt++;

	swapRange(a, nLft, nBegRgt-1);
	swapRange(a, nBegRgt, nRgt);
}