Force buckle 875 Coco, who likes bananas

This is a mathematical problem with a dichotomy . Through this problem, I learned a little skill about accuracy ：

It can be seen from the meaning of the question , If we want to eat $piles[i]$ Time spent , Can pass $\lceil \frac{piles[i]}{mid}\rceil$ To calculate , among $mid$ Is the median of two points . But this will involve the problem of accuracy when the number is very large , Such as $piles[i]=1000000000$,$mid=999999986$, At this point, use the above formula to calculate the result is 2, But actually it should be 3 That's right .

resolvent
because $piles[i]>0$,$mid>0$, So the original formula is equivalent to $\lfloor \frac{piles[i]+mid-1}{mid}\rfloor$, This is equivalent to... Without affecting the results $piles[i]$ It's magnified , If the number gets bigger, the result will get bigger , Therefore, the problem of accuracy is avoided . Specifically, you can substitute the above numbers into the experience .

The idea of dichotomy is very good , Don't go into details . Go straight to the code ：

class Solution {
    
    public int minEatingSpeed(int[] piles, int h) {
    
        Arrays.sort(piles);
        if (h == piles.length) {
    
            return piles[piles.length-1];
        }
        
        int l = 1;
        int r = piles[piles.length-1];
        while (l < r) {
    
            int mid = (l + r) >> 1;
            int count = 0;
            for (int i=0;i<piles.length;i++) {
    
                count += (piles[i] + mid - 1) / mid;
            }
            if  (count > h) {
    
                l = mid + 1;
            }
            else r = mid;
        }
        return l;
    }
}