lintcode254- Drop Eggs- easy

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.

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.

Example

Given n = 10, return 4.
Given n = 100, return 14.

用公式 1 + 2 + 3 + ... + n > 楼层，找到这个n即可。（小心sum用long来做，避免算的时候溢出。）

原理见这一页：http://datagenetics.com/blog/july22012/index.html

思想是让第一个鸡蛋落每一层，最后的worstcase都尽量接近即可优化。接近方法就是第一个鸡蛋每多存活一次，就多消耗一次次数，接下来就跳少一格留机会给第二个鸡蛋。

public class Solution {
    /*
     * @param n: An integer
     * @return: The sum of a and b
     */
    public int dropEggs(int n) {
        // write your code here
        long sum = 0;
        int i = 0;

        while (sum < n){
            i++;
            sum += i;
        }
        return i;
    }
}