以leetcode某题为背景:(所有代码均可在leetcode测试通过)

快速排序:

  1. 先按照字典序构造排序规则;(小写字母>大写字母>1-9);
  2. 分治策略实现,递归+while循环进行元素交换。
class Solution:
    def minNumber(self, nums: List[int]) -> str:
        def compare(x, y):
            # 转为字符串按照字典序进行比较;也就是直接比大小
            a = str(x) + str(y)
            b = str(y) + str(x)
            if a < b:
                return 0
            else:
                return 1


        def quick_sort(nums, left, right):
            if left > right:
                return
            low = left
            high = right
            priority = nums[left]
            while left < right:
                # 这里的循环,如果nums[right] 比 priority小才能继续进行,所以大数被转移到左边
                while left < right and compare(nums[right], priority):
                    right = right - 1
                nums[right], nums[left] = priority, nums[right]  # 因为这个时候的left是空出来的
                while left < right and compare(priority, nums[left]):
                    left = left + 1
                nums[left], nums[right] = priority, nums[left]

            quick_sort(nums, low, left - 1)
            quick_sort(nums, left + 1, high)

            return nums

        nums = quick_sort(nums, 0, len(nums) - 1)
        nums=''.join(str(num) for num in nums)
        return nums

归并排序:

class Solution:
    # 典型的分治法
    def minNumber(self, nums: List[int]) -> str:
        def compare(x, y):
            # 转为字符串按照字典序进行比较;也就是直接比大小
            a = str(x) + str(y)
            b = str(y) + str(x)
            if a < b:
                return 1
            else:
                return 0


        def merge(left, right):
            result = []
            i = j = 0
            while i < len(left) and j < len(right):
                if compare(left[i], right[j]):
                    result.append(left[i])
                    i = i + 1
                else:
                    result.append(right[j])
                    j = j + 1
            if i < len(left):
                result.extend(left[i:])
            if j < len(right):
                result.extend(right[j:])
            print(result)
            return result


        # 归并排序先拆分,再对不可拆分的最小项进行排序
        def merge_sort(nums):
            # 边界条件,当只有一个元素时 return
            if len(nums) <= 1:
                return nums
            mid = int(len(nums) / 2)
            # 包含边界
            left = merge_sort(nums[:mid])
            right = merge_sort(nums[mid:])
            return merge(left, right)
        nums=merge_sort(nums)
        nums=''.join([str(num) for num in nums])
        return nums

堆排序:

堆排序的流程:

  1. 首先是建堆,大顶堆是升序,小顶堆是降序(因为建好堆之后,每次需要把堆顶也就是最小的元素移动到数组的尾部);
    1. 选取i=length/2处的元素开始调整;之后继续向前直到i=0;
    2. 调整时可能会导致下面的数据混乱,所以要继续向下调整;
  2. 排序的过程,每次需要把堆顶元素和未标记的叶子结点进行交换;然后重新调整;
class Solution:
    def minNumber(self, nums: list[int]):
        def compare(x, y):
            # 转为字符串按照字典序进行比较;也就是直接比大小
            a = str(x) + str(y)
            b = str(y) + str(x)
            if a < b:
                return 0
            else:
                return 1

        # 调整堆内元素,每次调用只完成一次调整
        def adjust(nums, parent, length):
            temp = nums[parent]
            child = 2 * parent + 1
            while child <= length:
                if child < length and compare(nums[child], nums[child + 1]):
                    child = child + 1

                if not compare(temp, nums[child]):
                    break
                # 只会引起下级的不平衡,不会混乱
                nums[parent], nums[child] = nums[child], nums[parent]
                parent = child
                child = child * 2 + 1

        # 构造初始大顶堆
        border = len(nums) // 2 - 1
        for i in range(border, -1, -1):
            adjust(nums, i, len(nums) - 1)

        # 升序排列,大顶堆
        for i in range(len(nums) - 1, 0, -1):
            nums[0], nums[i] = nums[i], nums[0]
            adjust(nums, 0, i - 1)
        print(nums)
        result=''
        while nums:
            result+=str(nums.pop())
        return result

版权声明:本文为Andrehao原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明。
原文链接:https://blog.csdn.net/Andrehao/article/details/117416440