coursera算法公开课练习题Interview（2）

1. Question: 3-SUM in quadratic time. Design an algorithm for the 3-SUM problem that takes time proportional to n^2 in the worst case. You may assume that you can sort the n integers in time proportional to n^2 or better.
   

   问题描述: 这里的问题是想要一个增长数量级是平方级别或更好的3-sum算法

   解决方案: 增加一个哈希表来实现，先使用嵌套的2层循环，然后在哈希表中查询-(a[i] + a[j]), 当然要在数据结构里面加入flag数组来表示每个数是否作为嵌套循环里面的基底使用过了，如果哈希表中寻找到这个值且flag为false，那么count加1。

2. Question: Search in a bitonic array. An array is bitonic if it is comprised of an increasing sequence of integers followed immediately by a decreasing sequence of integers. Write a program that, given a bitonic array of n distinct integer values, determines whether a given integer is in the array.

   • Standard version: Use ∼3lgn compares in the worst case.
   • Signing bonus: Use ∼2lgn compares in the worst case (and prove that no algorithm can guarantee to perform fewer than ∼2lgn compares in the worst case).

   问题描述: 这是关于双调数组搜索的问题，所谓的双调数组的特点是，数组元素先递增，再递减，类似于一个峰的形式。然后标准实现方法的最坏情况是~3lgn的比较次数，然后最好的实现方法的最坏情况比较次数是关于~2lgn的。

   解决方案: 这里的标准解决方法很好实现，先使用二分法的变式找到数组的最大值，然后左右分别进行二分查找即可，很容易得到~3lgn的比较次数。对于更有效率的查找方式，这里我选择不查找最大值。具体的d递归实现方式描述如下：

   1. 先找到mid，将其与key比较，如果相等则返回true，否则进行下一步；
   2. 将mid值和左右两个值进行比较，判断mid在山峰的左边还是右边亦或是峰顶；
   3. 如果是峰顶，直接采取左右两边二分查找；如果是峰左，那么判定key值与mid值的相对大小，如果key值大于mid值，那么可以知道key值位于mid的右边，直接递归bitonicSearch(mid+1, hi, key)即可；如果key值小于mid值，那么需要对左边进行二分查找以及对右边进行递归查找；对于mid位于峰右的情况是同理的，具体Java代码见底部。

3. Question: Egg drop. Suppose that you have an n-story building (with floors 1 through n) and plenty of eggs. An egg breaks if it is dropped from floor T or higher and does not break otherwise. Your goal is to devise a strategy to determine the value of T given the following limitations on the number of eggs and tosses:
   Version 0: 1 egg, ≤T tosses.
   Version 1: ∼1lgn eggs and ∼1lgn tosses.
   Version 2: ∼lgT eggs and ∼2lgT tosses.
   Version 3: 2 eggs and ∼2√n tosses.
   Version 4: 2 eggs and ≤c√T tosses for some fixed constant c.

   问题描述: 这个是关于爬楼扔鸡蛋确定安全楼层的问题。

   解决方案: 其实这个就是解决两个子问题进行递归的问题，分别是鸡蛋碎了和鸡蛋没碎两种情况，而这两种情况也分别对应了确定的不安全楼层和安全楼层。

public class Bitonic {    private int[] bArray;    private int size;    public Bitonic(final int[] arr) {        size = arr.length;        bArray = new int[size];        for (int i = 0; i < size; i++) {            bArray[i] = arr[i];        }    }    public boolean search(int low, int hi, int key) {        int mid = low + (hi - low) / 2;        if (bArray[mid] == key) {            return true;        }        if (mid == 0) {            return search(mid + 1, hi, key);        } else if (mid == size - 1) {            return search(low, mid - 1, key);        } else {            if (bArray[mid - 1] < bArray[mid] && bArray[mid] > bArray[mid + 1]) {   // mid                return bArray[mid] < key ? false :((BinarySearch.indexOf(bArray, low, mid - 1, key) != -1)                        || (BinarySearch.indexOfReverse(bArray, mid + 1, hi, key) != -1));            } else if (bArray[mid - 1] < bArray[mid]) {    //left                return bArray[mid] < key ? search(mid + 1, hi, key) :                    ((BinarySearch.indexOf(bArray, low, mid - 1, key) != -1) || search(mid + 1, hi, key));            } else {    //right                return bArray[mid] < key ? search(low, mid - 1, key) :                    ((BinarySearch.indexOf(bArray, mid + 1, hi, key) != -1) || search(low, mid - 1, key));            }        }     }