调整数组 使得数组中负数排在前面正数排在后面，并且正数和负数的相对位置不变 (递归思路）

假设一整型数组存在若干正数和负数，现在通过某种算法使得该数组的所有负数在正数的左边，且保证负数和正数间元素相对位置不变。时空复杂度要求分别为：o(n)和o(1)

解题思路:
假设X代表正数，Y代表负数，下标代表正负数的序列位置，数组大小为arraySize
设有一数组 [X1, Y1, Y2, X2, Y3, Y4, X3, Y5, X4, X5] (可以理解为 [1, -1, -2, 2, -3, -4, 3, -5, 4, 5]
解题目标是把数组变为[Y1, Y2, Y3, Y4, Y5, X1, X2, X3, X4, X5] 而且时间复杂度为O(N), 空间复杂度为O(1)

step1:
扫描数组，可以指导数组中正数的数目 positiveNum

step2:
根据step1中得知的正数数目把数组分为两部分，第一部分大小为 arraySize-positiveNum，第二部分大小为positiveNum
[X1, Y1, Y2, X2 Y3] [Y4, X3, Y5, X4, X5]
并且对所得的两个子数组继续采用相同的划分方式进行递归操作
([X1, Y1, Y2] [X2, Y3]) ([Y4, X3][Y5, X4, X5])
(([X1, Y1][Y2]) ([X2][Y3])) (([Y4][X3]) ([Y5][X4, X5]))
(([X1][Y1]) [Y2] ) ([X2][Y3])) (([Y4][X3])
([Y5][X4, X5]))
这样划分的思想是:
当前一数组和后一数组分别完成排序后，前一数组中的正数等于后一数组中的负数，因为后一数组的大小就是原来整个数组中正数的大小
举个例子:
假设在数组 [X1,
Y1, Y2, X2, Y3, Y4, X3, Y5, X4, X5] 中，数组的划分为，后一数组的大小为正数的大小，为5
[X1,
Y1, Y2, X2 Y3] [Y4, X3, Y5, X4, X5]
当前一数组和后一数组分别完成排序后为
[Y1,
Y2, Y3, X1, X2] [Y4, Y5, X3, X4, X5]

因为正数的数目为5，前一数组中正数的数目(X1, X2)等于后一数组中的负数数目(Y4, Y5)

step3:
对于每两个不能再划分的子数组，进行归并。归并规则为把前一数组中的正数(如果有的话)和后一数组中的负数的位置进行一一互换
(([Y1, X1] [Y2]) [Y3, X2]) ([Y4, X3] [Y5, X4, X5])
([Y1, Y2, X1][Y3, X2]) ([Y4, Y5, X3, X4, X5])
([Y1, Y2, Y3, X1, X2]) ([Y4, Y5, X3, X4, X5])
([Y1, Y2, Y3, Y4, Y5, X1, X2, X3, X4, X5])

算法代码：

int getPositiveNum(int *pArray, int arraySize)
{
int count = 0;

if (pArray == NULL || arraySize <= 0)
{
return 0;
}

for (int idx = 0; idx < arraySize; idx++)
{
if (pArray[idx] > 0)
{
count++;
}
}
return count;
}

void swap(int *a, int *b)
{
int tmp;

tmp = *a;
*a  = *b;
*b  = tmp;
}

int sortArray(int *pArray, int arraySize)
{
int positiveNum = getPositiveNum(pArray, arraySize);

// if there is no positive number in the array, just return
if (positiveNum == 0)
{
return 0;
}

// if there is no negative number in the array, just return
if (positiveNum == arraySize)
{
return positiveNum;
}

/*
*  here what we going to do is
*  setup two sub arrays,
*  so the first sub array only have the nagtive number
*  the secnod sub array only have the positive number
*/

// here we can decide the size of each of the two sub array
int firstArraySize = arraySize-positiveNum;
int secnodArraySize = positiveNum;

/*
*  here we recursively separate the two sub array
*  after separation, we need to know how many positive numbers in the fist array
*  (that is also how many negative numbers in the secnod array)
*/
int positiveNumInFirstArray = sortArray(pArray, firstArraySize);
sortArray(&pArray[arraySize-positiveNum], secnodArraySize);

/*
*  here we presume that in each of the two sub array
*  negative numbers are in the front position, and they are in the right position
*  positive numbers are in the rear position, and they are in the right position
*  so, what we need to do is
*  swap each of the positive number in the first array and the negatinve number in the second array
*/
if (positiveNumInFirstArray > 0)
{
for (int pos = arraySize-positiveNum-positiveNumInFirstArray; pos <= arraySize-positiveNum-1; pos++)
{
swap(&pArray[pos], &pArray[pos + positiveNumInFirstArray]);
}
}

return positiveNum;
}

测试代码：

void printArray(int *pArray, int arraySize)
{
for (int idx = 0; idx < arraySize; idx++)
{
printf("%d ", pArray[idx]);
}
}

int _tmain(int argc, _TCHAR* argv[])
{
int intArray[] = {1,-1,-2,2,-3,-4,3,-5,4,5,-6,6,-7,-8,7,-9,8,9,10,-10};
int sizeArray = sizeof(intArray)/sizeof(intArray[0]);

printf("Before sorting the array------------\n");
printArray(intArray, sizeArray);

sortArray(intArray, sizeArray);

printf("After sorting the array-------------\n");
printArray(intArray, sizeArray);
}