二分查找

69. x 的平方根

 

 

class Solution {
public:
    int mySqrt(int x) {
        if (x == 0) return 0;  //特殊判断：0的平方根是0
       //0: 0; 1: 1; 2: 1; 3: 1
       //除上述4个数字外，其余数字x的平方根必小于等于其一半,即x/2
       //由此，找到了二分查找的边界
       int left = 1, right = x/2;
       while(left < right){
           long long mid = left + (right-left+1)/2;
           long long s = mid * mid;
           if(s > x){
               //找小于等于x的最大的s
               right = mid-1;
           }
           else{
               left = mid;
           }
        }
        return left;
    }
};

 

287. 寻找重复数

 

 

class Solution {
public:
    int findDuplicate(vector<int>& nums) {
        int left = 1, right = nums.size()-1;  //right为数组中数字的最大值
        while(left < right){
            int mid = left + (right-left)/2;
            int cnt = 0;
            for(int num: nums){
                if(num <= mid) cnt++;
            }
            if(cnt > mid){
                //说明重复的数字就在[left, mid]之间
                right = mid;
            }
            else{
                //说明重复的数字不在[left, mid]之间
                left = mid + 1;
            }
        }
        return left;
    }
};

875. 爱吃香蕉的珂珂

 

 

 

 

二分搜索的上界是数组piles中最大的数字，下界是1

class Solution {
public:
    int hours(vector<int>& piles, int k){
        //cout<<"k: "<<k<<endl;
        int ans = 0;
        for(int pile: piles){
            if(pile <= k){
                ans += 1;
            }
            else{
                //ans += ceil((double)pile / k);   //向上取整
                ans += (pile + k - 1)/k;
            }
        }
        //cout<<"ans: "<<ans<<endl;
        return ans;
    }
    int minEatingSpeed(vector<int>& piles, int H) {
        //随着最小速度k增加，可以在H小时内吃掉所有香蕉的H减少
        int left = 1;
        int right = *max_element(piles.begin(), piles.end());
        while(left < right){
            int mid = left + (right-left)/2;
            int hour = hours(piles, mid);
            if(hour > H){
                //花费的时间比H大，则速度k增加
                left = mid + 1;
            }
            else{
                right = mid;
            }
        }
        return left;
    }
};

 

LCP 12. 小张刷题计划