1. 寻找两个正序数组的中位数

2. 两数相除

3. 快速幂

4. 搜索旋转排序数组

5. 数组中的逆序对

6. 在排序数组中查找元素的第一个和最后一个位置

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        return {find(nums,target,true),find(nums,target,false)};
    }
    int find(vector<int>& nums, int target, bool minType) {
        int left = 0, right = nums.size() - 1;
        int ans = -1;//默认为-1
        while (left <= right) {//当搜索区间位于两指针之间
            int mid = left + (right - left) / 2;//二分查找
            if (nums[mid] == target) {
                ans = mid;
                if (minType) right = mid - 1;//找左边界（注意没写反）
                 else left = mid + 1;}//找右边界（注意没写反）
             else if (target < nums[mid]) right = mid - 1;//移动右指针到二分位置
             else left = mid + 1;//左指针移到二分位置
            
        }
        return ans;
    }
};

7. 找出第k小的数对距离

class Solution {
public:
    int smallestDistancePair(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());//排序
        int n = nums.size();
        int left = 0, right = nums.back() - nums.front();
        while (left <= right) {
            int mid = (left + right) / 2;
            int cnt = 0;//计算所有距离小于等于mid的数对数目cnt
            for (int j = 0; j < n; j++) {
            int i = lower_bound(nums.begin(), nums.begin() + j, nums[j] - mid) - nums.begin();
            cnt += j - i;
            }
            if (cnt >= k) right = mid - 1;
            else left = mid + 1;
        }
        return left;
    }
};

8. 爱吃香蕉的珂珂

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {
        long sum =0;int max_ = 0;
        for(int x:piles){
            sum+=x;
            max_ = max(max_,x);}
        int min_ = ceil(sum*1.0/h);//向上取整
        while(min_<max_){
            int times = 0;
            int mid = (max_+min_)/2;
            for(int x:piles)
                times = times + ceil(x*1.0/mid);
            if(times==h) max_ = mid; //找左边界动右边
            else if(times<h) max_ = mid; 
            else if(times>h) min_ = mid+1;//找左边界
        }
        return min_;
    }
};