875. Koko Eating Bananas

最新推荐文章于 2025-07-29 14:24:32 发布

转载最新推荐文章于 2025-07-29 14:24:32 发布 · 62 阅读

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原文链接：http://www.cnblogs.com/ruruozhenhao/p/9940100.html

文章标签：

#c/c++ #runtime

本文探讨了Koko在限定时间内吃完所有香蕉堆的最小速度问题，通过二分查找算法确定最佳吃香蕉速度，确保在规定时间内完成。示例展示了不同情况下的最优解，并附带了C++代码实现。

Koko loves to eat bananas. There are N piles of bananas, the i-th 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 won't eat any more bananas during this hour.

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

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

Note:

1 <= piles.length <= 10^4
piles.length <= H <= 10^9
1 <= piles[i] <= 10^9

Approach #1:

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int H) {
        int len = piles.size();
       if (len == 1) return 1;
        long long l = 0, r = pow(10, 9);
        while (l < r) {         // not l <= r
            long long mid = l + (r - l) / 2;
            int time = 0;
            for (int i = 0; i < len; ++i) 
                time += (piles[i] - 1) / mid + 1; // better than time += (piles[i] % mid) == 0 ? piles[i] / mid : piles[i] / mid + 1;
            if (time > H) l = mid + 1;
            else r = mid;       // mot r = mid - 1
        }
        return l;
    }
};

Runtime: 68 ms, faster than 38.03% of C++ online submissions for Koko Eating Bananas.

Analysis:

Pay attention to the boundary condition.

转载于:https://www.cnblogs.com/ruruozhenhao/p/9940100.html