C经典算法之二分查找法
1.根据key查找所在数组的位置
#include <stdio.h>/* key = 9; 1 2 3 4 5 6 7 8 arr 3， 4， 5， 7， 9 ， 11， 21， 23 low = 1 mid = (low + high)/2 = 4 high = 8; one arr[mid] = 7 < 9; so low = mid + 1 = 5; high = 8; mid = (low + high)/2 = 6 two arr[mid] = 11 > 9 so low = 5 , high = mid - 1 = 5 mid = 5; arr[mid] = 9 == key if(key = 10) low = mid + 1 > high */int main(int argc, const char * argv[]){ int findByHalf(int arr[], int len, int key); int arr[] = {3, 4 , 5, 7, 9 , 11, 21, 23}; int len = sizeof(arr)/sizeof(int); int index = findByHalf(arr, len, 88); printf("index = %d\n", index); return 0;}int findByHalf(int arr[], int len, int key){ int low = 0; int high = len - 1; int mid ; while(low <= high){ mid = (low + high) / 2; //右边查找 if (key > arr[mid]) { low = mid + 1; //左边查找 }else if (key > arr[mid]) { high = mid - 1; }else{ return mid; } } return -1;}
2.插入一个数，得到其所在数组的位置
#include <stdio.h>/* key = 9; 1 2 3 4 5 6 7 8 arr 3， 4， 5， 7， 9 ， 11， 21， 23 low = 1 mid = (low + high)/2 = 4 high = 8; one arr[mid] = 7 < 9; so low = mid + 1 = 5; high = 8; mid = (low + high)/2 = 6 two arr[mid] = 11 > 9 so low = 5 , high = mid - 1 = 5 mid = 5; arr[mid] = 9 == key if(key = 10) low = mid + 1 > high */int main(int argc, const char * argv[]){ int findByHalf(int arr[], int len, int key); int arr[] = {3, 4 , 5, 7, 9 , 11, 21, 23}; int len = sizeof(arr)/sizeof(int); int index = findByHalf(arr, len, 88); printf("index = %d\n", index); return 0;}int insertByHalf(int arr[], int len, int key){ int low = 0; int high = len - 1; int mid ; while(low <= high){ mid = (low + high) / 2; //右边查找 if (key > arr[mid]) { low = mid + 1; //左边查找 }else if (key > arr[mid]) { high = mid - 1; }else{ //如果arr[mid] == key //就把key插入到这个数的后面 return mid + 1; } } //如果low > high 说明 key > arr[mid]; //就把key插入到low对应的 这个数的位置 return low;}
如有疑问请留言或者到本站社区交流讨论，感谢阅读，希望能帮助到大家，谢谢大家对本站的支持！
获致幸福的不二法门是珍视你所拥有的遗忘你所没有的。
