二分法练习题目

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二分查找法

1. 二分查找的数组是有序的

2. 二分查找能够有效的减少有序数组的查询次数

3. 题目

利用二分查找法，查找某个数，如果存在则输出这个数，如果不存在，输出比它大的最小的数，输出比它小的最大的数，如果该数不在这个数组的范围内，就会报错

#include<stdio.h>#include<stdlib.h>void init(int array[],int num);int binaryFind(int array[],int num,int findNum);//06void main(){int ret = 1;int a[10] ={0};init(a,sizeof(a)/sizeof(*a));int findNum = 0;printf("输入你想要查找的数：");scanf("%d",&findNum);    ret = binaryFind(a,sizeof(a)/sizeof(*a),findNum);if (ret == -1){printf(" error: %d  can't find the number %d",ret,findNum);}else{printf("the label is %d",ret+1);}system("pause");}//数组输入和排序void init(int array[],int num){   int temp = 0;   //数组的录入   for (int i = 0;i< num;i++)   {   printf("请输入%d个数",i+1);   scanf("%d",&array[i]);   }   //冒泡排序   for(int i = 0;i< num;i++)   {      for(int j = 0;j<num-i-1;j++)  {  if(array[j]>array[j+1])  {  temp = array[j];  array[j] = array[j+1];  array[j+1] = temp;  }  }   }}//二分查找int binaryFind(int array[],int num,int findNum){int ret = -2;if (array == NULL){return ret = -1;}//二分查找主要算法int begin = 0;int end   = num -1;int nowLable = 0;if (findNum < array[0] || findNum > array[num -1]){return ret = -1;}while(begin <= end){nowLable = (begin + end)/2;if (array[nowLable]== findNum){return nowLable;}if (array[nowLable]>findNum){end = nowLable - 1;}else{        begin = nowLable + 1;}}if (array[nowLable] > findNum){printf("the biger number is %d\n",array[nowLable]);printf("the lower number is %d\n",array[nowLable -1]);}else{printf("the biger number is %d\n",array[nowLable+1]);printf("the lower number is %d\n",array[nowLable]);}return ret;}

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