heap sort complexity

Ext.: http://www.cppblog.com/converse/archive/2008/09/28/63008.html The merge sort algorithm (mergesort) divides a sequence into two sub-sequences of the same size, then sorts the two sub-sequences, and finally merges the two subsequence sequences into an ordered sequence. In the process of synthesis, the general realization all need to open up a space with the same size as the original sequence, For merging operations, an example of a merge

million, but if you remove the duplicates, no more than 3 million. The higher the repetition of a query string, the more users are queried for it, the more popular it is, please count the hottest 10 query strings, which requires no more than 1G of memory.Solution: Although there are 10 million query, but because of the high repetition, so in fact only 3 million of the query, each query255byte, (300w*255b Hash statistics: This batch of massive data preprocessing (maintain a key for the

A set of randomly sorted alphanumeric numbers. Write an algorithm with a time complexity of O (n) so that the letters are sorted from small to large. Description: The letters are case-sensitive, the same letters, and the sort guard lowercase rows are in front. For example: r,b,b,b,w,w,b,r,b,w sort to:b,b,b,b,b,r,r,w,w,w#includeToday is too late, first put out the

spacewhile (Iif (r[i].key{R1[k]=r[i];i++;k++;}else//elements in the 2nd paragraph are put into R1{R1[K]=R[J];j++;k++;}while (I{R1[k]=r[i];i++;k++;}while (J{R1[K]=R[J];j++;k++;}for (k=0,i=low;iR[I]=R1[K];Free (R1);}An algorithm for merging a tripvoid Merpass (RecType r[],int length,int N)//The entire sort sequence is merged{int i;for (i=0;i+2*length-1Merge (r,i,i+length-1,i+2*length-1);if (i+length-1Merge (r,i,i+length-1,n-1); Merge these two sub-tabl

heap) #include 4, the time complexity of the heap and the complexity of spacethe best, worst, and average time complexity of the heap is O (nlogn).the space complexity of the

//Take the larger child nodes in the left and right child nodes of I - if(J -j + +; the - if(A[i] -Swap (A[i], a[j]); - Else + Break; -} +} A at //From right to do, from bottom to top - voidHeapsort (intA[],intN -{ - ///First step: Establish Dagen (I is the first non-leaf node, n is the number of nodes) - for(inti = N/2; I >= 1; i--) -{ inHeapadjust (A, I, N); -} to + for(inti = n; I >= 1; i--) -{ the //Second step: Swap

Both of these operations are recursive implementations, Hanoi thought.#include #include using namespace STD;voidReverse Stackint>s)//Reverse Stack{if(s.size () = =0)return;intA=s.top (); S.pop ();if(s.size () = =0) {S.push (a);return; } reverse (s);intB=s.top (); S.pop (); Reverse (s); S.push (a); Reverse (s); S.push (b);}voidSort ( Stackint>s)//stack top magnification element{if(s.size () = =0)return;intA=s.top (); S.pop ();if(s.size () = =0) {S.push (a);return; }

The implementation of the sort heap and the use of heaps to sort table of Contents A few basic definitions Heapy: Adjusting the heap Build: Build a heap Insert: Inserting a new element Removetop: Remove the top of the

For example , array a:[0:5,1:2,2:4,3:3,4:1]The results are as follows: [1, 2, 3, 4, 5], but sometimes there is a need to keep the pre-order position in a single array, as in the preceding example: [4, 1, 3, 2, 0], so a separate array ordering process is implemented using heap sorting.The code is as follows:functionArraykeys (arr) {vari = 0, Len=arr.length, Keys= []; while(I Len) {Keys.push (i++); } returnkeys; } 　　 //Determine if the vari

/** (max) heap down adjustment algorithm * * Note: array implementation of the heap, the nth node of the left child's index value is (2n+1), the right child's index is (2n+2). * where n is an array subscript index value, and the 1th number in the array corresponds to n 0. * * Parameter Description: * A--the array to be sorted * Start--The starting position of the node being lowered (typically 0, indicating

Code that was previously written Public void bubblesort (int arr[]) { for (int i = 0, len = arr.length; i -1; i++) { for (int j = 0; J ) {if(arr[j + 1] Arr[j]) + 1); }}}have never considered the best case of time complexity why is O (n), a sudden look, it is not scientific ah?Until I read the following code, it dawned on me. Public voidBubblesort (intarr[]) { BooleanDidswap; for(inti = 0, len = arr.lengt

("----------++++++++++++---------------------/n"); } Sorting Core Algorithm Description One, treat the first record as an ordered subsequence Second, sequentially, from the second record, insert the ordered subsequence by comparison. Client Test Code int main (int argc, char* argv[]){int i=0;Calogrithm demo;Demo.setlength (5); Demo.setmemkey (1, 6);Demo.setmemkey (2, 20);Demo.setmemkey (3, 15);Demo.setmemkey (4, 7);Demo.setmemkey (5, 3);Demo.cout ();Demo. Insertsort ();Demo.cout ();return 0;} T

Four, (1) heap sorting The first time to listen to heap sorting is in 107lab listen to brother, but did not say how to achieve. At that time just looked at the data structure of the two-tree, but also thought to be through the pointer to establish a two-tree way to achieve, feel very difficult. In fact, the heap sort

back, and the corresponding nextpos ++ You can write it in one sentence: for (int c = A[i]; c When I = 0, nextpos is changed to {1, 1, 1}, but a [0] is changed to 2. So we need to reverse the for loop so that a [0] is still 0 for (int c = 2; c >= A[i]; c--) A[nextPos[c]++] = c; However, this loop does not run correctly, because when C = 2, a [0] is changed to 2, c -- void sortColors(int A[], int n) { int nextPos[3] = {0}; for (int i = 0; i

code for this step is simple:/* Heap Sorting, Idea: * First maxheapbuild, make the root node the maximum value * then the root node and the last node's value is exchanged * then the last node is disconnected (maxsize-=1) * and then maxheapify, Because this time the root node of the left and right sub-tree is still a large heap, so only need maxheapify and do not need maxheapbuild * So from n-1 start cycle

Heap Sort1. 堆： 1. 一种完全二叉树。 2. 每个结点的值都大于或等于其左右子结点的值，大顶堆。 3. 小顶堆同理。2. 是简单选择排序的一种改进：把每次比较的结果用堆来保存起来。3. 堆排序（大顶堆）： 1. 将待排序列构造成一个大顶堆。 2. 将堆顶和待排序列最后一个元素交换，也就是保存起来。 3. 将剩余的序列（去除最后一个元素）重新构造成一个堆。 4. 重复23 。4. 待排序列构造初始大顶堆： 1. 设序列长度length，已经构造好最初的完全二叉树，无序。 2. 从最下层最右边的非叶子结点开始向左向上。 3. 二叉树的性质：根节点从序号1开始，设某个结点的序号为k，则其左子树的序号是2k，右子树的序号是2k+1。最下层最右边的非叶子结点就是length/2取整。 4. 从第【length/2取整】个结点开始，向根节点（序号为1）开始 逐个调整每个子树的三个结点的顺序，让其成为大顶堆。 5. 交换之后可

First, the topic understandingTitle: Array al[0,mid-1] and al[mid,num-1] are each ordered, two sub-ordered segments of the array al[0,num-1] to merge, get al[0,num-1] overall order. Requires a space complexity of O (1). Note: The al[i] element is supported by the ' The first chapter of the data structure tells the order table merging, but then it is merged into the new table, the condition is while (iSecond, the implementation of the algorithmSet two

only need to maxheapify and not need maxheapbuild * This will start from n-1 to 2. */ void Maxheapsort (int a[], int maxSize, int tmp[]) { Maxheapbuild (A, maxSize); //n For (int i = maxSize; i > 0;) { //cycle n-1 times Swap (A, a + i); Tmp[i] = A[i]; i--; if (i = = 0) { Tmp[0] = a[0]; } Else { Maxheapindex (A, 0, i); //logn executed n-2 times. } } For (int i = 0; I A[i] = Tmp[i]; } } iv. Complexity