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875. Koko Eating Bananas
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: 4Example 2:
Input: piles = [30,11,23,4,20], H = 5
Output: 30Example 3:
Input: piles = [30,11,23,4,20], H = 6
Output: 23Note:
-
1 ⇐ piles.length ⇐ 10^4 -
piles.length ⇐ H ⇐ 10^9 -
1 ⇐ piles[i] ⇐ 10^9
思路分析
二分查找，确定左边界
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/**
* 二分查找，确定左边界
*
* @author D瓜哥 · https://www.diguage.com
* @since 2024-08-15 20:40:08
*/
public int minEatingSpeed(int[] piles, int h) {
int left = 1, right = 0;
for (int pile : piles) {
right = Math.max(right, pile);
}
while (left < right) {
int mid = left + (right - left) / 2;
int hours = hours(piles, mid);
if (hours == h) {
// 搜索左侧边界，需要收缩右侧边界
right = mid;
} else if (hours < h) {
// 由于消耗时间小于指定时间，
// 那么就需要降低速度，来提高消耗时间。
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
/**
* 以 k 速度，吃完所有香蕉需要的时间
*/
private int hours(int[] piles, int k) {
int result = 0;
for (int pile : piles) {
result += pile / k;
if (pile % k != 0) {
result++;
}
}
return result;
}