drop eggs

There is a building of n floors. If an egg drops from the k th floor or above, it will break. If it's dropped from any floor below, it will not break.

You're given two eggs, Find k while minimize the number of drops for the worst case. Return the number of drops in the worst case.

Example

Example 1:

Input: 100
Output: 14

Example 2:

Input: 10
Output: 4

Clarification

For n = 10, a naive way to find k is drop egg from 1st floor, 2nd floor ... kth floor. But in this worst case (k = 10), you have to drop 10 times.

Notice that you have two eggs, so you can drop at 4th, 7th & 9th floor, in the worst case (for example, k = 9) you have to drop 4 times.

思路：目标就是找到一种策略，使得无论哪一楼破裂，我们找到那一层所用到的数目都是一样的，也就是怎么将N怎么分成几段，然后每次找的次数都是一样的。

例如：总数是10，那么我们从4，7，9测试，所得到的次数都是4次。

4破裂 1，2，3， 总共1+3 = 4次；

7破裂，4没破裂，5，6，总共也是4次

9破裂，4，7没破裂，8，总共也是4次。

如果分的原则就是第x , 第x+ (x-1) ,第X+(x-1) +(x-2)  ....

也就是x +(x-1) +(x-2) +...+2+1 >= n

我们就是要找到这个x是多少第一次使得公式成立。

https://www.youtube.com/watch?v=NGtt7GJ1uiM

public class Solution {
    /**
     * @param n: An integer
     * @return: An integer
     */
     // x + (x-1) + (x-2) + (x-3) + .. + 1 >=n;
    public int dropEggs(int n) {
        if(n <= 0) return 0;
        
        long index = 1;
        while(index*(index+1) / 2 < n){
            index = index*2;
        }
        
        long start = 1;
        long end = index;
        
        // find first x which let x*(x+1) / 2 >= n;
        while(start + 1 < end) {
            long mid = start +(end - start) / 2;
            if(mid*(mid+1) / 2 > n){
                end = mid;
            } else {
                start = mid;
            }
        }
        
        if(start*( start+1) / 2 >= n){
            return (int)start;
        }
        return (int)end;
    }
}