如题。二分查找简单快捷,但也可以多种变化,其中按返回值可以将其分为如题中的三种。简单实现+测试如下。
STL中的lower_bound和upper_bound实现类型功能,参考带源码:https://blog.csdn.net/qq_36172505/article/details/77380559
使用lower_bound(begin,end,greater)可以实现在降序数组二分查找。
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// 对相同元素,返回任意位置
// 对不存在的查找,返回任意位置(或-1)
int bs1(int target, vector<int> nums) {
int L = 0, R = nums.size() - 1;
while (L <= R) {
int m = L + (R - L) / 2;
if (nums[m] < target) {
L = m + 1;
}
else if (nums[m] > target) {
R = m - 1;
}
else {
return m;
}
}
return -1;
}
// 对相同元素,返回下标最小位置
// 对不存在的查找,返回不大于target的最大位置
int bs2(int target, vector<int> nums) {
int L = 0, R = nums.size() - 1;
while (L <= R) {
int m = L + (R - L) / 2;
if (nums[m] < target) {
L = m + 1;
}
else if (nums[m] >= target) {
R = m - 1;
}
}
if (nums[R+1] == target)
++R;
return R;
}
// 对相同元素,返回下标最大位置
// 对不存在的查找,返回不小于target的最小位置
int bs3(int target, vector<int> nums) {
int L = 0, R = nums.size() - 1;
while (L <= R) {
int m = L + (R - L) / 2;
if (nums[m] <= target) {
L = m + 1;
}
else if (nums[m] > target) {
R = m - 1;
}
}
if (nums[R] != target)
++R;
return R;
}
void Print(const vector<int>& nums) {
for (int i = 0, size = nums.size(); i < size; ++i)
cout << nums[i] << " ";
cout << endl;
}
void PrintN(int index, const vector<int>& nums) {
for (int i = 0; i < index; ++i)
cout << " ";
cout << index << endl;
}
int main() {
vector<int> nums1{1,2,4,5,5,5,5,6};//奇数
vector<int> nums2{ 1,2,2,4,5,5,5,5,5,6,8,8 };//偶数
cout << "一般查找:" << endl;
//成功查找
cout << "成功查找:查找5" << endl;
Print(nums1);
PrintN(bs1(5,nums1), nums1);
Print(nums2);
PrintN(bs1(5, nums2), nums2);
//失败查找
cout << "失败查找:查找3" << endl;
Print(nums1);
PrintN(bs1(3, nums1), nums1);
Print(nums2);
PrintN(bs1(3, nums2), nums2);
cout << endl;
cout << "返回不大于目标的最大下标---查找:" << endl;
//成功查找
cout << "成功查找:查找5" << endl;
Print(nums1);
PrintN(bs2(5, nums1), nums1);
Print(nums2);
PrintN(bs2(5, nums2), nums2);
//失败查找
cout << "失败查找:查找3" << endl;
Print(nums1);
PrintN(bs2(3, nums1), nums1);
Print(nums2);
PrintN(bs2(3, nums2), nums2);
cout << endl;
cout << "返回不小于目标的最小下标---查找:" << endl;
//成功查找
cout << "成功查找:查找5" << endl;
Print(nums1);
PrintN(bs3(5, nums1), nums1);
Print(nums2);
PrintN(bs3(5, nums2), nums2);
//失败查找
cout << "失败查找:查找3" << endl;
Print(nums1);
PrintN(bs3(3, nums1), nums1);
Print(nums2);
PrintN(bs3(3, nums2), nums2);
cout << endl;
return 0;
}