題目：
輸入一個長度為N的正整數陣列piles代表N堆香蕉，piles[i]代表第i堆香蕉的數量，現在要在H小時內吃完之這些香蕉，請求假設吃香蕉的速度為k，而且每小時最多吃一堆香蕉，如果吃不下的話留下到下一個小時再吃，如果吃完後這堆後，如果還有胃口，也要等待下一個小時才能吃下一堆。
求最小的k
分析：
1）一堆香蕉N，每小時吃K個，則一堆需要多久呢？
如果N%k=0，則需要N/k 小時
如果N%k！=0.則需要N/k + 1小時
2) 此問題最直接的解法是，k從1遍歷到max(piles[i])，因為再大已經沒有意義，最多一小時只能吃一堆，針對每個值計算是否可以在H小時吃完，計算第一個滿足的k，則是最小的k
/**
* 有n堆食物，在H小時內吃完，每小時吃多少個食物，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從1開始找到max速度不夠快，可以採用二分查詢的方法，比如如果max=100，使用50，發現能在H小時內吃完，則k最小值，應該在1~50內，否則在50~100內，再次縮小到25，12，6 比直接從1遍歷要快很多
/**
* 有n堆食物，在H小時內吃完，每小時吃多少個食物，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;
//二分查詢
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;
}
}