Leetcode 875. Koko Eating Bananas (Binary search 经典题)-优快云博客

文章讲述了Koko面临香蕉堆，要在守卫回来前吃完所有香蕉，要求找到在给定时间内吃得最慢但能吃完的最小速度。使用二分查找算法解决此问题，给出几个示例和时间复杂度分析。

Koko Eating Bananas
Medium
Koko loves to eat bananas. There are n piles of bananas, the ith pile has piles[i] bananas. The guards have gone and will come back in h hours.

Koko can decide her bananas-per-hour eating speed of k. Each hour, she chooses some pile of bananas and eats k bananas from that pile. If the pile has less than k bananas, she eats all of them instead and will not eat any more bananas during this hour.

Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.

Return the minimum integer k such that she can eat all the bananas within h hours.

Example 1:

Input: piles = [3,6,7,11], h = 8
Output: 4
Example 2:

Input: piles = [30,11,23,4,20], h = 5
Output: 30
Example 3:

Input: piles = [30,11,23,4,20], h = 6
Output: 23

Constraints:

1 <= piles.length <= 104
piles.length <= h <= 109
1 <= piles[i] <= 109

解法1：经典binary search。注意即使某个解已经满足条件，比如时间刚好等于h，也不一定是最优解，因为可能比它小的速度也可以满足时间刚好等于h。
比如：
Input: piles = [3,6,7,11], h = 8
速度为5满足h=8，但最优速度为4。

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h) {
        long long n = piles.size();
        long long start = 1, end = LLONG_MAX;
        while (start + 1 < end) {
            long long mid = start + (end - start) / 2;
            int hour = eatingHours(piles, mid) ;
            if (hour <= h) {
                end = mid;
            } else if (hour > h) {
                start = mid;
            } else {
                return mid;
            }
        }
        if (eatingHours(piles, start) <= h) return start;
        return end;
    }
private:
    // return # of hours Koko used to finish all the bananas with speed k
    // the higher k, the less the hours needed
    long long eatingHours(vector<int> piles, long long k) {
        long long res = 0;
        long long n = piles.size();
        for (int i = 0; i < n; i++) {
            res += piles[i] / k + (piles[i] % k > 0 ? 1 : 0);
        }
        return res;
    }
};