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
Solution: Binary Search, 找到最小的K。possible里pile-1使得pile-1/k是小于1的。(不然如果pile为4,K = 4,那么time一次性加了2)如果possible为假,那么我们要找的最小的K肯定在mi之上,如果possible为真,我们就继续找最小的,让hi = mi。
class Solution { public: int minEatingSpeed(vector<int>& piles, int H) { int lo = 1, hi = pow(10,9); while(lo < hi){ int mi = lo + (hi - lo) / 2; if(!possible(piles, H, mi)) lo = mi + 1; else hi = mi; } return lo; } bool possible(vector<int>& piles, int H, int K){ int time = 0; for(auto pile : piles){ time += (pile - 1) / K + 1; } return time <= H; } };