问题
So i guess its because it just compares A[k] and A[k-1], and does the implementation in one sweep but its still not clear. Can someone explain better. Thanks
回答1:
This link shows a graphical representation of sorting algorithm with different types of data set. As you can see, when the data is sorted the algorithm complexity is reduced to N. Which is equivalent to the number of elements as inputs.
The link provided gives a clear picture of how its more efficient.
回答2:
You answered your own question: For a nearly sorted array, insertion sort will only need a handful of O(n) passes to complete. Contrast that to a divide and conquer sorting algorithm like merge sort, which takes O(n*lgn). For any non trivial value of n, a divide and conquer algorithm will need many O(n) passes, even if the array be almost completely sorted, whereas insertion sort might only require a few.
回答3:
Insertion sort is a faster and more improved sorting algorithm than selection sort. In selection sort the algorithm iterates through all of the data through every pass whether it is already sorted or not. However, insertion sort works differently, instead of iterating through all of the data after every pass the algorithm only traverses the data it needs to until the segment that is being sorted is sorted. Again there are two loops that are required by insertion sort and therefore two main variables, which in this case are named 'i' and 'j'. Variables 'i' and 'j' begin on the same index after every pass of the first loop, the second loop only executes if variable 'j' is greater then index 0 AND arr[j] < arr[j - 1]. In other words, if 'j' hasn't reached the end of the data AND the value of the index where 'j' is at is smaller than the value of the index to the left of 'j', finally 'j' is decremented. As long as these two conditions are met in the second loop it will keep executing, this is what sets insertion sort apart from selection sort. Only the data that needs to be sorted is sorted.
回答4:
The general goal of a sorting algorithm is to minimize the number of comparisons. Sorting algorithms have a lower bound and an upper bound on the number of comparisons( n log n worst-case for merge and heap sorts, n log n average case for quick sort). In the most general case, you'd go with an algorithm that happens to have the best average or best worst-case number of comparisons. However, when you know something about the data (e.g., the array is already sorted, or almost sorted), you can exploit the fact that insertion sort's lower bound is far lower than the "n log n" sorts.
For example, if you have an array [1,2,3,4,5,6,7,9] and you need to insert 8 into it, you can either insert it at the end, and sort the array using a vanilla n log n sort (which will do about 28 comparisons (roughly) to sort the data to [1,2,3,4,5,6,7,8,9]). However, insertion sort lets you insert the 8 at the right position in only about 8 comparisons.
来源：https://stackoverflow.com/questions/35884216/why-insertion-sort-is-best-algorithm-for-sorted-or-nearly-sorted-array