谷歌面试题：输入是两个整数数组，他们任意两个数的和又可以组成一个数组，求这个和中前k个数怎么做？

“假设两个整数数组为A和B，各有N个元素，任意两个数的和组成的数组C有N^2个元素。
那么可以把这些和看成N个有序数列：
A[1]+B[1] <= A[1]+B[2] <= A[1]+B[3] <=…
A[2]+B[1] <= A[2]+B[2] <= A[2]+B[3] <=…
…
A
+B[1] <= A
+B[2] <= A
+B[3] <=…
问题转变成，在这N^2个有序数列里，找到前k小的元素”

//得到首中尾3个数的中位数
int getMidian(int array[], int low, int high) {
int mid = low + ((high - low) >> 1);
if (array[mid] > array[high]) {
swap(array[mid], array[high]);
}
if (array[low] > array[high]) {
swap(array[low], array[high]);
}
if (array[low] < array[mid]) {
swap(array[mid], array[low]);
}

return array[low];
}

int kth_elem(int array[], int low, int high, int th) {
int pivot = getMidian(array, low, high);

int lowTmp = low;
int highTmp = high;

while (lowTmp < highTmp) {
while (lowTmp < highTmp && array[highTmp] > pivot) {
highTmp--;
}
array[lowTmp] = array[highTmp];
while (lowTmp < highTmp && array[lowTmp] < pivot) {
lowTmp++;
}
array[highTmp] = array[lowTmp];
}
array[lowTmp] = pivot;

if (th - 1 == lowTmp) {
return array[lowTmp];
} else if (th - 1 < lowTmp) {
return kth_elem(array, low, lowTmp - 1, th);
} else {
return kth_elem(array, lowTmp + 1, high, th);
}
}