# LeetCode875. Koko Eating Bananas（二分）

## 2019 年 06 月 13 日 • 阅读: 1128 • LeetCode • 阅读设置

### LeetCode875. Koko Eating Bananas（二分）

• Medium
• Accepted：13,211
• Submissions：28,647

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

### 链接

https://leetcode.com/problems/koko-eating-bananas

### 代码

• Runtime: 72 ms, faster than 38.20% of C++ online submissions for Koko Eating Bananas.
• Memory Usage: 10.3 MB, less than 47.57% of C++ online submissions for Koko Eating Bananas.
class Solution {
public:
int minEatingSpeed(vector<int>& piles, int H) {
int n = piles.size();
sort(piles.begin(), piles.end());
int l = 1, r = piles[n - 1];
while (l < r)
{
int mid = (l + r) >> 1;
if (isValid(piles, H, mid)) // mid符合条件，答案在[l, mid]中
r = mid;
else
l = mid + 1;
}
return l;
}

bool isValid(vector<int>& piles, int H, int x)
{
int sum = 0;
for (int i = 0; i < piles.size(); ++i)
sum += (piles[i] - 1) / x + 1; // 巧妙的处理，相当于 piles[i] / x + piles[i] % x;
return sum <= H;
}
};

The end.
2019年6月13日 星期四