External sort

External sorting

External storage , Data exchange between memory

The principle of external sorting

The size of each buffer should be equal to the size of the disk block

Multi channel balanced merging

There is only one buffer empty , It should be immediately filled with the next piece of the merged paragraph , Instead of directly 27 Output to output buffer

The merged results will be written to another space on the disk , And release the original space , Here, the merged results are moved to the original space in the view to demonstrate the beauty

Increase the number of ways to merge , Reduce the number of merging trips , This can reduce the number of times to read and write to the disk

Loser tree

Some books call the position of Tianjin rice in the loser tree the root node , Some people call the location of Sunwukong the root node

Now there are the following 8 A merge segment , Select the smallest data element each time , Every smallest element needs to be selected 7 Second comparison . To reduce the number of comparisons , We can build a loser tree first .
Each leaf node in the loser tree below corresponds to a merge segment

At these failed nodes ( That is, the grey node in the loser tree ) Usually, the merging segment from which the failed element is recorded , The following screenshot is for convenience , So it's the value of the element , In fact, the gray node should be used to store the merging segment from which the competing elements failed

The correct node records are as follows , That is, the gray node records which merging segment the loser comes from , The root node records the champion ( The final winner of the competition ) From which merging segment

Merge for the first time , Contrast 7 We found the smallest element from which merge segment . Next, according to the rules of merging and sorting , We also need to merge 1~8 Select the next smallest element in the . Next, we'll let the merging section 3 The next element of replaces 1 The original position of this element

Next, select the new smallest element from the remaining leaf nodes , We just need to let 6 Compare with the smallest element in the fourth merging segment



As long as the loser tree is constructed , Next, each time you select the smallest element , Just compare 3 Time , That is, the number of grey nodes of the loser tree

Suppose we have built a loser tree , Tree height is h(h Blue nodes are not included , It's a tree of grey and green nodes ). Is the first h There are $2^{h-1}$ Nodes
k The loser tree of the road merger will have k Leaf node . So there should be $k<=2^{h-1}$
Solution $h-1=\lceil {\log }_{2}k\rceil$ . and h-1 It just represents how many layers the branch node has . As mentioned before, the number of layers of a branch node needs to be compared , So with the loser tree , Choose the smallest element , Just compare keywords $\lceil {\log }_{2}k\rceil$ Time
If it is 1024 Road merging , Then the traditional method , It's going to happen every time 1023 Compare it to , The loser tree only needs comparison $\lceil {\log }_{2}1024\rceil$ namely 10 Second comparison

The array index is zero 1 The element of corresponds to the root node in the traditional sense , The array index is zero 0 Corresponding to the newly added small head ls[0]. Leaf nodes do not correspond to any data in the actual array . Logically , Each green leaf node corresponds to a merge segment , In fact, these leaf nodes are made up by our brain

substitution - Selection sort ( Generate initial merge segment )

The best merging tree