EPI：Sorting

最新推荐文章于 2025-09-28 03:01:35 发布

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本文探讨了针对不同情况下的高效排序算法，包括快速排序、插入排序、堆排序、计数排序、基于指针的间接排序等，并介绍了如何在变量长度排序中封装对象以顺利进行交换操作。此外，还涉及了基于二叉搜索树（BST）、映射（Map）和计数排序的实现，以及任务分配、最大并发事件数计算、区间并集、插入区间、最小覆盖点等问题的解决方法。

(1) Most efficient sorting algorithm for each situation

-- A large array whose entries are random numbers

Quicksort. considered one of the most efficient sorting, and is in-place

-- A small array of numbers

Insert sorting. stable and in-place

-- A large array of numbers that are already sorted

Heap sort.

-- A large collection of integers that are drawn from a small range

Counting sort

-- A large collection of numbers most of which are duplicates

BST with linked list for elements which have the same key

-- Stabilily is required

Merge sort, efficient and stable

(2) Variable length sort

思路：几乎所有的sorting都依赖于swapping操作，但是，如果交换对象的大小不一样，如果才能顺利进行交换呢？能不能将对象封装成大小一样的item呢？答案是Yes。Build an array P of pointers to the records. Then sort the pointers using the compare function on dereferenced pointers。 Finally, iterate through P writing the deference pointers

(3) Least distance sort.

思路：同上（Indirect sort）, sort references to the objects first and then apply the permutation that was applied the references in the end. In this way, each item is moved exactly correct destination exactly once.

(4) BST, Map, Counting sort

Map is implemented by the BST, and counting sort is entailed by the integer range.

(5) Task assignment

compute a 2-partition of A that has min max load

(6) Maximum number of events concurrently

思路：sort the endpoints of events, and count the maximum number of events taking place concurrently

(7) Union of intervals

思路：维护一个start和end

(8) Insert an interval

思路：类似于merge intervals

(9) Minimum points covering intervals

思路：find the task which ends earliest and remove those tasks which intersect with it. Repeat this process and continue.

(10) Stack sort (increasing)

思路：递归，拿出最顶段的item, insert it into the left sorted items in the stack.

if(S.empty()==false){
    T e=S.top();
    S.pop();
    sort(S);
    insert(S, e);
}

void insert(stack<int>& S, int e){
      if(S.empty() || S.top()>=e) 
              S.push(e);
      else{
              int f=S.top();
              S.pop();
              insert(S,e);
              S.push(f);
      }
}