两种算法一起使用
冒泡算法的时间复杂度是n的平方,二分法是log(m+n)
#include <iostream> #include <vector> #include <string> int main() { std::vector<int> nums{ 20,10,324,43,415,2,45,68,9,100,98}; int len = nums.size(); int low = 0, high = 0, temp, mid; if (len % 2 != 0) { //奇数 mid = (len - 1) / 2; } else { //偶数 mid = len / 2 - 1; } int n; //冒泡排序 for (int i = 0; i < len; i++) { for (int j = 0; j < len - i - 1; j++) { if (nums[j] > nums[j + 1]) { temp = nums[j]; nums[j] = nums[j + 1]; nums[j + 1] = temp; } } } for (int k = 0; k < len; k++) { std::cout << nums[k] << " "; } std::cout << std::endl; //二分法找到这个数的下标 std::cin >> n; if (nums[0] == n) { std::cout << "下标: 0" << std::endl; return 0; } if (nums[len-1] == n) { std::cout << "下标: " <<len-1<< std::endl; return 0; } while (1) { if (n == nums[mid]) { std::cout << "下标:" << mid << std::endl; break; } else if (n > nums[mid]) { low = nums[mid + 1]; if (low == n) { std::cout << "下标:" << mid + 1 << std::endl; break; } mid = (mid + len) / 2; } else { high = nums[mid - 1]; mid = (mid + 0) / 2; } } return 0; }