题目连接

解题思路

使用二分搜索，左边界为0，右边界为时间最大值。然后判断条件是看存不存在连续的花朵能够凑成我们的mid天数，能就让右边界变成mid

代码

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class Solution {
public:
    bool check(vector<int> bloomDay, int m, int k, int time){
        int bouque = 0,num = 0;
        for(auto day:bloomDay){
            if(day<=time){
                num++;
                if(num == k){
                    bouque++;
                    num=0;
                }
            }else{
                num = 0;
            }
        }
        return bouque>=m;
    }
    int minDays(vector<int>& bloomDay, int m, int k) {
        int l = 0, r = 0;
        for(auto day:bloomDay){
            r = max(day,r);
        }
        r++;
        int flag = 0;
        while(l < r){
            int mid = (l + r) >> 1;
            if(check(bloomDay,m,k,mid)){
                flag = 1;
                r = mid;
            }else{
                l = mid + 1;
            }
        }
        return flag?r:-1;
    }
};