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  • b树的实现

    花了蛮长时间实现的b树插入操作。有时间再实现其他操作。

    
    #include <stdio.h>
    #include <stdlib.h>
    #define M 5
    
    enum KeyStatus { Duplicate,SearchFailure,Success,InsertIt,LessKeys };
    struct node {
            int n; /* n < M No. of keys in node will always less than order of B tree */
            int keys[M-1]; /*array of keys*/
            struct node *p[M]; /* (n+1 pointers will be in use) */
    }*root=NULL;
    
    
    int pushIn(struct node* currentNode, int key , struct node* toInsertNode){
        int i=0;
    
    
        for(i=currentNode->n -1;i>=-1;i--){
            if(key < currentNode->keys[i]){
                currentNode->keys[i+1]=currentNode->keys[i];
          currentNode->p[i+2]=currentNode->p[i+1];
            }
            else{
                currentNode->keys[i+1]=key;
          currentNode->p[i+2]= toInsertNode;
                break;
            }
    
        }
      currentNode->n++;
        return 0;
    }
    
    int searchNode(struct node* currentNode, int key, int *downNodeIndex){
        int i=0;
      for(i=currentNode->n -1;i>=0;i--){
        if(key== currentNode->keys[i]){
          return Success;
        }else if (key < currentNode->keys[i]){
          if( i>0 ){
            continue;
          }else {
            *downNodeIndex=0;
          }
        }else{
          *downNodeIndex=i+1;
          break;
        }
      }
      return SearchFailure;
    }
    
    int splitNode(struct node* currentNode, int toInsertkey, struct node* toInsertNode, int *toLiftKey, struct node** toLiftNode ){
      struct node* rightSplittedNode = calloc(sizeof(struct node),1);
      int i=0;
      int leftMedian=((M-1)/2 - 1);
      int rightMedian=(M-1)/2;
      if(toInsertkey > currentNode->keys[rightMedian]){
        for(i=M-1;i>=rightMedian+2;i--){
          pushIn(rightSplittedNode,currentNode->keys[i-1],currentNode->p[i]);
          currentNode->n--;
        }
        pushIn(rightSplittedNode,toInsertkey, toInsertNode);
        *toLiftKey=currentNode->keys[currentNode->n-1];
        rightSplittedNode->p[0]=currentNode->p[currentNode->n];
        *toLiftNode=rightSplittedNode;
        currentNode->n--;
      }
      else {
        for(i=M-1;i>=rightMedian+1;i--){
          pushIn(rightSplittedNode,currentNode->keys[i-1],currentNode->p[i]);
          currentNode->n--;
        }
        pushIn(currentNode,toInsertkey, toInsertNode);
        *toLiftKey=currentNode->keys[currentNode->n-1];
        rightSplittedNode->p[0]=currentNode->p[currentNode->n];
        *toLiftNode=rightSplittedNode;
        currentNode->n--;
      }
    
      return 0;
    
    }
    
    
    int pushDown(struct node* currentNode, int toInsertKey, int* toLiftKey, struct node** toLiftNode){
    
      int downNodeIndex;
      struct  node* toInsertNode;
      int rc;
    
      if( NULL == currentNode){
        *toLiftNode=NULL;
        (*toLiftKey)=toInsertKey;
        return InsertIt;
      }
    
      if( Success == searchNode(currentNode,toInsertKey, &downNodeIndex) ){
        return Duplicate;
      }
    
      rc=pushDown(currentNode->p[downNodeIndex], toInsertKey, toLiftKey, toLiftNode);
    
      if(InsertIt !=rc){
        return rc;
      }
    
      if( currentNode->n < M-1 ){
        pushIn(currentNode,*toLiftKey,*toLiftNode);
        return Success;
      }else {
        toInsertKey=(*toLiftKey);
        toInsertNode=*toLiftNode;
        splitNode(currentNode,toInsertKey,toInsertNode,toLiftKey,toLiftNode);
        printf("toLiftNode %d
    ", toLiftNode);
      }
    
      return rc;
    
    }
    
    
    int insert(struct node * currentNode, int key){
      struct node* toLiftNode;
      struct node* newRoot;
      int toLiftKey;
      int rc;
      toLiftNode=NULL;
      rc=pushDown(currentNode,key, &toLiftKey,&toLiftNode);
    
      if(InsertIt == rc ){
        newRoot =  calloc( sizeof(struct node),1);
        newRoot->n=1;
        newRoot->keys[0]=toLiftKey;
        newRoot->p[0]=root;
        newRoot->p[1]=toLiftNode;
        root=newRoot;
        rc=Success;
      }
      return rc;
    }
    
    void display(struct node *ptr, int blanks , int level)
    {
        if (ptr)
        {
            int i;
              printf("level:%d nodeAddr:%ld   ",level, ptr);
            for(i=1; i<=blanks; i++)
                printf(" ");
            for (i=0; i < ptr->n; i++)
                printf("%d ",ptr->keys[i]);
            printf("
    ");
            for (i=0; i <= ptr->n; i++)
                display(ptr->p[i], blanks+10,level+1);
        }/*End of if*/
    }/*End of display()*/
    
    int main(int argc, char const *argv[]) {
      insert(root,1*10);
      insert(root,2*10);
      insert(root,3*10);
      insert(root,4*10);
      display(root,0,0);
      insert(root,9);
      insert(root,5*10);
      insert(root,6*10);
      insert(root,7*10);
      display(root,0,0);
      insert(root,8*10);
      insert(root,8);
      insert(root,7);
      insert(root,6);
      insert(root,5);
      insert(root,4);
      insert(root,3);
      insert(root,2);
      insert(root,41);
      insert(root,42);
      insert(root,43);
      insert(root,44);
    
      display(root,0,0);
      return 0;
    }
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  • 原文地址:https://www.cnblogs.com/youge-OneSQL/p/7899484.html
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