我知道中位数算法的中位数（我将表示为MoM）是一个高常数因子O（N）算法。它找到k组的中位数（通常为5），并将它们用作下一个迭代的集合来查找中位数。找到它之后的枢轴将在原始集合的3 / 10n和7 / 10n之间，其中n是查找一个中间基本情况所需的迭代次数。
当我为MoM运行此代码时，我一直遇到分段错误，但我不确定原因。我调试了它，并认为问题在于我正在调用medianOfMedian(medians, 0, medians.size()-1, medians.size()/2);。但是，我认为这在逻辑上是合理的，因为我们应该通过调用自己来递归地找到中位数。也许我的基本情况不正确？在YogiBearian在youtube（斯坦福教授，链接：https://www.youtube.com/watch?v=YU1HfMiJzwg）的教程中，他没有说明任何额外的基础案例来处理MoM中递归的O（N / 5）操作。
注意：根据建议，我添加了一个基本案例，并使用矢量使用.at（）函数。
static const int GROUP_SIZE = 5;
/* Helper function for m of m. This function divides the array into chunks of 5
* and finds the median of each group and puts it into a vector to return.
* The last group will be sorted and the median will be found despite its uneven size.
*/
vector<int> findMedians(vector<int>& vec, int start, int end){
vector<int> medians;
for(int i = start; i <= end; i+= GROUP_SIZE){
std::sort(vec.begin()+i, min(vec.begin()+i+GROUP_SIZE, vec.end()));
medians.push_back(vec.at(min(i + (GROUP_SIZE/2), (i + end)/2)));
}
return medians;
}
/* Job is to partition the array into chunks of 5(subject to change via const)
* And then find the median of them. Do this recursively using select as well.
*/
int medianOfMedian(vector<int>& vec, int start, int end, int k){
/* Acquire the medians of the 5-groups */
vector<int> medians = findMedians(vec, start, end);
/* Find the median of this */
int pivotVal;
if(medians.size() == 1)
pivotVal = medians.at(0);
else
pivotVal = medianOfMedian(medians, 0, medians.size()-1, medians.size()/2);
/* Stealing a page from select() ... */
int pivot = partitionHelper(vec, pivotVal, start, end);
cout << "After pivoting with the value " << pivot << " we get : " << endl;
for(int i = start; i < end; i++){
cout << vec.at(i) << ", ";
}
cout << "\n\n" << endl;
usleep(10000);
int length = pivot - start + 1;
if(k < length){
return medianOfMedian(vec, k, start, pivot-1);
}
else if(k == length){
return vec[k];
}
else{
return medianOfMedian(vec, k-length, pivot+1, end);
}
}
以下是我为这两个函数编写的一些单元测试。希望他们有所帮助。
vector<int> initialize(int size, int mod){
int arr[size];
for(int i = 0; i < size; i++){
arr[i] = rand() % mod;
}
vector<int> vec(arr, arr+size);
return vec;
}
/* Unit test for findMedians */
void testFindMedians(){
const int SIZE = 36;
const int MOD = 20;
vector<int> vec = initialize(SIZE, MOD);
for(int i = 0; i < SIZE; i++){
cout << vec[i] << ", ";
}
cout << "\n\n" << endl;
vector<int> medians = findMedians(vec, 0, SIZE-1);
cout << "The 5-sorted version: " << endl;
for(int i = 0; i < SIZE; i++){
cout << vec[i] << ", ";
}
cout << "\n\n" << endl;
cout << "The medians extracted: " << endl;
for(int i = 0; i < medians.size(); i++){
cout << medians[i] << ", ";
}
cout << "\n\n" << endl;
}
/* Unit test for medianOfMedian */
void testMedianOfMedian(){
const int SIZE = 30;
const int MOD = 70;
vector<int> vec = initialize(SIZE, MOD);
cout << "Given array : " << endl;
for(int i = 0; i < SIZE; i++){
cout << vec[i] << ", ";
}
cout << "\n\n" << endl;
int median = medianOfMedian(vec, 0, vec.size()-1, vec.size()/2);
cout << "\n\nThe median is : " << median << endl;
cout << "As opposed to sorting and then showing the median... : " << endl;
std::sort(vec.begin(), vec.end());
cout << "sorted array : " << endl;
for(int i = 0; i < SIZE; i++){
if(i == SIZE/2)
cout << "**";
cout << vec[i] << ", ";
}
cout << "Median : " << vec[SIZE/2] << endl;
}
Given array :
7, 49, 23, 48, 20, 62, 44, 8, 43, 29, 20, 65, 42, 62, 7, 33, 37, 39, 60, 52, 53, 19, 29, 7, 50, 3, 69, 58, 56, 65,
After pivoting with the value 5 we get :
23, 29, 39, 42, 43,
After pivoting with the value 0 we get :
39,
Segmentation Fault: 11
在分段错误之前，似乎没事。我确信我的分区功能也是可行的（这是leetcode问题的实现之一）。
免责声明：这不是一个家庭作业问题，而是我在leetcode问题集中使用quickSelect后对算法的好奇心。
如果我提出的问题需要更多详细说明MVCE，请告诉我，谢谢！
编辑：我发现递归分区方案在我的代码中是错误的。正如Pradhan指出的那样 - 我不知何故有空向量导致开始和结束分别为0和-1，导致我从调用它的无限循环中产生分段错误。仍然试图弄清楚这一部分。
答案 0 :(得分：3)
MoM 始终调用自身（计算pivot），从而展示无限递归。这违反了递归算法的“主要指令”：在某些时候，问题“小”到不需要递归调用。
答案 1 :(得分：2)
在Scott的暗示的帮助下，我能够正确实现这个中位数算法的正确实现。我修复它并意识到我的主要想法是正确的，但有一些错误：
我的基本情况应该是大小为＆lt; = 5的子向量。
关于最后一个数字（变量末尾），在这种情况下是否应该被认为是包含在内，或者上限是否小于，有一些小的微妙之处。在下面的这个实现中，我的上限小于定义。
这是下面的内容。我也接受了斯科特的回答 - 谢谢斯科特！
/* In case someone wants to pass in the pivValue, I broke partition into 2 pieces.
*/
int pivot(vector<int>& vec, int pivot, int start, int end){
/* Now we need to go into the array with a starting left and right value. */
int left = start, right = end-1;
while(left < right){
/* Increase the left and the right values until inappropriate value comes */
while(vec.at(left) < pivot && left <= right) left++;
while(vec.at(right) > pivot && right >= left) right--;
/* In case of duplicate values, we must take care of this special case. */
if(left >= right) break;
else if(vec.at(left) == vec.at(right)){ left++; continue; }
/* Do the normal swapping */
int temp = vec.at(left);
vec.at(left) = vec.at(right);
vec.at(right) = temp;
}
return right;
}
/* Returns the k-th element of this array. */
int MoM(vector<int>& vec, int k, int start, int end){
/* Start by base case: Sort if less than 10 size
* E.x.: Size = 9, 9 - 0 = 9.
*/
if(end-start < 10){
sort(vec.begin()+start, vec.begin()+end);
return vec.at(k);
}
vector<int> medians;
/* Now sort every consecutive 5 */
for(int i = start; i < end; i+=5){
if(end - i < 10){
sort(vec.begin()+i, vec.begin()+end);
medians.push_back(vec.at((i+end)/2));
}
else{
sort(vec.begin()+i, vec.begin()+i+5);
medians.push_back(vec.at(i+2));
}
}
int median = MoM(medians, medians.size()/2, 0, medians.size());
/* use the median to pivot around */
int piv = pivot(vec, median, start, end);
int length = piv - start+1;
if(k < length){
return MoM(vec, k, start, piv);
}
else if(k > length){
return MoM(vec, k-length, piv+1, end);
}
else
return vec[k];
}