LeetCode-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 Hhours.

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

题解：

第一次超时，不可顺序查找

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int H) {
      int n = piles.size();
      int mink = 0;
        for (int k = 1; ; k++) {
          double sum = 0;
          for (int i = 0; i < n; i++) {
            int idx = piles[i] / k;
            sum += idx;
            if (piles[i] % k != 0) {
              sum++;
            }
          }
          if (sum <= H) {
            mink = k;
            break;
          }
        }
      return mink;
    }
};

改用二分查找AC：

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int H) {
      int n = piles.size();
      sort(piles.begin(), piles.end());
      int left = 0, right = piles[n - 1];
      if (n == 1) {
        return (piles[0] + H - 1 )/H;
      }
      if (n == H) {
        return right;
      }
      while (left <= right) {
        int k = (left + right) / 2;
        int sum = 0;
        for (int i = 0; i < n; i++) {
          int idx = piles[i] / k;
          sum += idx;
          if (piles[i] % k != 0) {
            sum++;
          }
        }
        if (sum <= H) {
          right = k - 1;
        }
        else {
          left = k + 1;
        }
      }
      return left;
    }
};