Counting sort-优快云博客

本文详细介绍了计数排序算法的工作原理及其实现过程。通过两个不同的代码片段展示了如何处理非负整数排序及包含负数的情况，并解释了为什么计数排序的时间复杂度为O(n)。

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The running time of counting sort is O(n), and usually, the running time of sorting algorithms will be either O(n^2) or O(n lgn), the reason why counting sort is O(n) is that it doesn't have the comparison.In fact, the sorting is hidden from line 19 to line 27.

class CountingSortAlgorithm { public void CountingSort(int[] arrayA) { int max = arrayA[0]; for (int i = 1; i < arrayA.Length; i++) { if (arrayA[i] > max) { max = arrayA[i]; } } int[] arrayC = new int[max + 1]; int[] arrayB = new int[arrayA.Length]; for (int i = 0; i < arrayA.Length; i++) { arrayC[arrayA[i]] += 1; } for (int i = 1; i < arrayC.Length; i++) { arrayC[i] += arrayC[i - 1]; } for (int i = arrayA.Length - 1; i >= 0; i--) { arrayB[arrayC[arrayA[i]] - 1] = arrayA[i]; arrayC[arrayA[i]]--; } for (int i = 0; i < arrayA.Length; i++) { arrayA[i] = arrayB[i]; } } }

The code segment above is only suitable for the case in which all the elements in the array are non-negative. If the array

has negative element, we can simply 'shift' the element, such as the minimum element will be considered as 0. I use 'span'

to shift the element.

class CountingSortAlgorithm { public void CountingSort(int[] arrayA) { int max = arrayA[0]; int min = arrayA[0]; for (int i = 1; i < arrayA.Length; i++) { if (arrayA[i] > max) { max = arrayA[i]; } if (arrayA[i] < min) { min = arrayA[i]; } } int span = -min; int[] arrayC = new int[max - min + 1]; int[] arrayB = new int[arrayA.Length]; for (int i = 0; i < arrayA.Length; i++) { arrayC[arrayA[i] + span] += 1; } for (int i = 1; i < arrayC.Length; i++) { arrayC[i] += arrayC[i - 1]; } for (int i = arrayA.Length - 1; i >= 0; i--) { arrayB[arrayC[arrayA[i] + span] - 1] = arrayA[i]; arrayC[arrayA[i] + span]--; } for (int i = 0; i < arrayA.Length; i++) { arrayA[i] = arrayB[i]; } } }