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: 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

题目理解：

有若干堆苹果，每小时可以吃某一堆的n个，如果这一堆不足n个，那么只能吃掉这一堆里剩下的苹果，必须在H小时内吃完，问n的最小值是多少

解题思路：

采用逐渐尝试不同大小的n的方法。因为吃每一堆都至少需要一小时，因此n最大为苹果最多的那一堆得苹果数目，最少是1，因此我们可以采用二分法，不断尝试，找到最小的符合要求的n值

class Solution {
    public int minEatingSpeed(int[] piles, int H) {
        int len = piles.length, left = 0, right = 0;
        for(int num : piles) {
        	right = Math.max(right, num);
        }
        while(left < right) {
        	int mid = left + (right - left) / 2;
        	if(judge(piles, H, mid))
        		right = mid;
        	else
        		left = mid + 1;
        }
        while(!judge(piles, H, left))
        	left++;
        while(judge(piles, H, left - 1))
        	left--;
        return left;
    }
    
    public boolean judge(int[] piles, int H, int speed) {
        if(speed == 0)
    		return false;
    	int time = 0;
    	for(int num : piles) {
    		time += num / speed;
    		if(num % speed != 0)
    			time++;
    	}
    	return time <= H;
    }
}

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