快速排序和归并排序都使用了分治思想. 分治算法通常都用递归来实现python
分治: 分而治之, 将一个大问题不断的分解为小问题来解决, 小的问题解决了, 大的问题也就解决了.算法
思想: 将原数组不断分解为先后两部分, 直到每一个数组内只有一个元素, 而后不断进行排序合并, 最后合并为一个有序数组数组
O(logn)
次, 每次都须要对n个元素排序, 因此须要O(nlogn)
# coding:utf-8 def merge(left, right): res = [] while left and right: # 此处决定了排序是否稳定. 须要保证针对相等的元素排序后按照出现的前后顺序进行排列 if left[0] < right[0]: res.append(left.pop(0)) else: res.append(right.pop(0)) if left: res.extend(left) if right: res.extend(right) return res def merge_sort(nums): length = len(nums) if length <= 1: return nums middle = int(length / 2) left = merge_sort(nums[:middle]) right = merge_sort(nums[middle:]) return merge(left, right) if __name__ == "__main__": nums = [4, 3, 6, 9, 7, 0, 1, 9, 3] assert merge_sort(nums) == [0, 1, 3, 3, 4, 6, 7, 9, 9]
使用了分治思想. 以数组中的一个数key为基准, 把小于key的数放到左边, 把大于key的数放到右边, 而后使用一样的方法做用于key两边的区间数据结构
O(n**2)
# coding:utf-8 """ 空间复杂度: O(n) """ def quick_sort(nums): if len(nums) <= 1: return nums key = nums.pop() # 不考虑空间消耗 less, over = [], [] for i in nums: if i < key: less.append(i) else: over.append(i) return quick_sort(less) + [key] + quick_sort(over) if __name__ == "__main__": nums_1 = [4, 3, 6, 9, 7, 0, 1, 9, 3] assert quick_sort(nums_1) == [0, 1, 3, 3, 4, 6, 7, 9, 9]
# coding:utf-8 """ 空间复杂度: O(1) """ def partition(nums, low, high): key_index = high key = nums[key_index] while low < high: while low < high and nums[low] <= key: low += 1 while low < high and nums[high] >= key: high -= 1 nums[low], nums[high] = nums[high], nums[low] nums[low], nums[key_index] = nums[key_index], nums[low] return low def interval(nums, low, high): if low < high: new_index = partition(nums, low, high) interval(nums, 0, new_index - 1) interval(nums, new_index + 1, high) return nums def quick_sort(nums): res = interval(nums, 0, len(nums) - 1) return res if __name__ == "__main__": nums_2 = [4, 3, 6, 9, 7, 0, 1, 9, 3] assert quick_sort(nums_2) == [0, 1, 3, 3, 4, 6, 7, 9, 9]
O(n**2)
快速排序中最坏状况是分区后一个分区是空, 另外一个分区全满, 这种通常是key的选择不当致使的, 好比一个有序数组选择了第一个或最后一个元素为key, 能够采用如下方法优化app