lintcode254- Drop Eggs- easy

最新推荐文章于 2020-04-01 21:33:49 发布

转载最新推荐文章于 2020-04-01 21:33:49 发布 · 152 阅读

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原文链接：http://www.cnblogs.com/jasminemzy/p/7590228.html

本文探讨了利用两枚鸡蛋确定一栋楼中会使鸡蛋摔碎的最低楼层的问题。通过优化策略，文章提供了一种有效的方法来最小化最坏情况下的尝试次数，并给出了具体的实现代码。

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;
    }
}

转载于:https://www.cnblogs.com/jasminemzy/p/7590228.html