二分搜索算法-吃香蕉问题

题目：

      输入一个长度为N的正整数数组piles代表N堆香蕉，piles[i]代表第i堆香蕉的数量，现在要在H小时内吃完之这些香蕉，请求假设吃香蕉的速度为k，而且每小时最多吃一堆香蕉，如果吃不下的话留下到下一个小时再吃，如果吃完后这堆后，如果还有胃口，也要等待下一个小时才能吃下一堆。

求最小的k

分析：

1）一堆香蕉N，每小时吃K个，则一堆需要多久呢？

如果N%k=0，则需要N/k 小时

如果N%k！=0.则需要N/k + 1小时

2) 此问题最直接的解法是，k从1遍历到max(piles[i])，因为再大已经没有意义，最多一小时只能吃一堆，针对每个值计算是否可以在H小时吃完，计算第一个满足的k，则是最小的k

/**
 * 有n堆食物，在H小时内吃完，每小时吃多少个食物，最小的值是多少
 */
public class MinEatingSpeed {


    public static void main(String[] args) {

        int[] piles = new int[5];
        int H = 5;
        piles[0] = 10;
        piles[1] = 20;
        piles[2] = 30;
        piles[3] = 40;
        piles[4] = 50;
        System.out.println(minEatingSpeed(piles,H));
 
    }


    public static int minEatingSpeed(int[] piles, int H) {
        int max = max(piles);
        for (int speed = 1; speed < max; speed++) {
            if(canFinish(piles,speed,H)){
                return speed;
            }
        }
        return max;
    }

    public static Boolean canFinish(int[] piles,int speed,int H){
        int count = 0;
        for (int pile : piles) {
            count = count + ((pile%speed)==0?pile/speed:(pile/speed) + 1);
//            System.out.println("pile=" + pile + " speed=" + speed + " (pile%speed)==0?pile/speed:(pile/speed) + 1=" + (count + (pile%speed)==0?pile/speed:(pile/speed) + 1));
//            count = count + (pile/speed) + ((pile%speed)>0?1:0);
        }
        return count<=H;
    }


    public static int max(int[] piles){
        int max=0;
        for (int pile : piles) {
            if(pile>max){
                max=pile;
            }
        }
        return max;
    }
}

　　3）k从1开始找到max速度不够快，可以采用二分查找的方法，比如如果max=100，使用50，发现能在H小时内吃完，则k最小值，应该在1~50内，否则在50~100内，再次缩小到25，12，6 比直接从1遍历要快很多

/**
 * 有n堆食物，在H小时内吃完，每小时吃多少个食物，最小的值是多少
 */
public class MinEatingSpeed {


    public static void main(String[] args) {

        int[] piles = new int[5];
        int H = 5;
        piles[0] = 10;
        piles[1] = 20;
        piles[2] = 30;
        piles[3] = 40;
        piles[4] = 50;
  
        //二分查找
        System.out.println(minEatingSpeed2(piles,H));

    }

    public static int minEatingSpeed2(int[] piles, int H) {

        int left = 1;
        int right = max(piles) + 1;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if(canFinish(piles,mid,H)){
                right = mid;
            }else{
                left = mid + 1;
            }
        }
        return left;
    }



    public static Boolean canFinish(int[] piles,int speed,int H){
        int count = 0;
        for (int pile : piles) {
            count = count + ((pile%speed)==0?pile/speed:(pile/speed) + 1);
//            System.out.println("pile=" + pile + " speed=" + speed + " (pile%speed)==0?pile/speed:(pile/speed) + 1=" + (count + (pile%speed)==0?pile/speed:(pile/speed) + 1));
//            count = count + (pile/speed) + ((pile%speed)>0?1:0);
        }
        return count<=H;
    }


    public static int max(int[] piles){
        int max=0;
        for (int pile : piles) {
            if(pile>max){
                max=pile;
            }
        }
        return max;
    }
}