Decrease-and-Conquer

                        第四章 Decrease-and-Conquer

 增量方法：自下而上的变化通常是以迭代方式实现；以问题的一个小地方开始。

The bottom-up variation is usually implemented iteratively, starting with a solution to the smallest instance of the problem; it is called sometimes the incremental approach.

Three major variations of decrease-and-conquer:

               Decrease by a constant
               Decrease by a constant factor
               Variable size decrease

4.1 Insertion Sort

      4.1.1 介绍

    如果我们已经找到了数组中n-2个元素的正确位置，那么第 A[n-1] 个元组放在哪呢？只需要在刚刚排好序的数组中，找到

一个位置插入这最后一个元素即可

Following the technique’s idea, we assume that the smaller problem of sorting the array A[0..n − 2] has already been solved to give us a sorted array of size n − 1: A[0]≤ . . . ≤ A[n − 2]. How can we take advantage of this solution to the smaller problem to get a solution to the original problem by taking into account the element A[n − 1]? Obviously, all we need is to find an appropriate position for A[n − 1] among the sorted elements and insert it there.

        这通常是通过从右到左扫描已排序的子阵列来完成的，直到遇到小于或等于[N-1]的第一个元素，在该元素后面插入一个[N-1]。

This is usually done by scanning the sorted subarray from right to left until the first element smaller than or equal to A[n − 1] is encountered to insert A[n − 1] right after that element. 

虽然插入排序显然基于递归思想，但是自下而上（即迭代）实现该算法更有效。

Though insertion sort is clearly based on a recursive idea, it is more efficient to implement this algorithm bottom up, i.e., iteratively

4.1.2 伪代码：

代码实现或其他详细解析请参考： https://blog.csdn.net/qq_36098284/article/details/98098798

4.1.3例子：

4.1.4  时间复杂度分析

该算法的关键是每次的比较过程，即 A[j] 和 v 的大小

The basic operation of the algorithm is the key comparison A[j] > v.

比较的次数依赖于输入的数据，最坏的情况是，当序列恰巧是降序，而我们希望得到升序的时候，从i = 1 开始，需要比较该数字其前面的所有数字，并且需要依次进行移动。最坏的情况下，比较的次数和选择排序相同。

T e number of key comparisons in this algorithm obviously depends on the nature of the input. In the worst case, A[j]> v is executed the largest number of times, i.e., for every j = i − 1, . . . , 0. Since v = A[i], it happens if and only if A[j]> A[i] for j = i − 1, . . . , 0.Thus, in the worst case, insertion sort makes exactly the same number of comparisons as selection sort

选择排序可以参考： https://blog.csdn.net/qq_36098284/article/details/97812203

因此当出现这种情况的时候，时间复杂是 O(n^2)

Thus, for the worst-case input, we get A[0]>A[1] (for i = 1), A[1]> A[2] (for i = 2), . . . , A[n − 2]> A[n − 1]

最好的情况是，序列是 升序，同时我们也想要升序，时间复杂度是 n - 1

In  the best case, the comparison A [j]> v is executed only once on every iteration of the outer loop. It happens if and only if A[i − 1]≤ A[i] for every

虽然平均效率是O(n^2) 使得，相对于其他的排序算法，还是比较脱出。

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4.2 Topological Sort

更新时间2019.8.9！！！

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