荷兰国旗三路划分

最新推荐文章于 2026-01-04 19:40:08 发布

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#算法 #leetcode

本文探讨了三种不同的算法问题：荷兰国旗问题的解决方案、使用快速排序实现的三路摆动序列调整，以及查找数组中第k大的元素的方法。通过实例代码展示了如何在Python中高效地解决这些经典问题，适合算法爱好者和开发者参考。

1.荷兰国旗问题`lc75

class Solution:
    def sortColors(self, nums: List[int]) -> None:
        n = len(nums)
        p0 = p1 = 0
        for i in range(n):
            if nums[i] == 1:
                nums[i], nums[p1] = nums[p1], nums[i]
                p1 += 1
            elif nums[i] == 0:
                nums[i], nums[p0] = nums[p0], nums[i]
                if p0 < p1:
                    nums[i], nums[p1] = nums[p1], nums[i]
                p0 += 1
                p1 += 1

2.三路 lc324摆动序列

class Solution:
    def wiggleSort(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        if n < 2: return nums
        mid = (0 + n - 1) // 2  # 中位数索引

        # 快速排序中的一次划分
        def partition(begin, end):
            left, right = begin, end
            while left < right:
                while left < right and nums[left] < nums[right]: right -= 1
                if left < right:
                    nums[left], nums[right] = nums[right], nums[left]
                    left += 1
                while left < right and nums[left] < nums[right]: left += 1
                if left < right:
                    nums[left], nums[right] = nums[right], nums[left]
                    right -= 1
            return left

        # 找到中位数对应的数值
        left, right = 0, n - 1
        while True:
            pivot = partition(left, right)
            if pivot == mid:
                break
            elif pivot > mid:
                right = pivot - 1
            else:
                left = pivot + 1

        # 三路划分(荷兰旗)
        midNum = nums[mid]
        left, curr, right = 0, 0, n - 1
        while curr < right:
            if nums[curr] < midNum:
                nums[left], nums[curr] = nums[curr], nums[left]
                left += 1
                curr += 1
            elif nums[curr] > midNum:
                nums[curr], nums[right] = nums[right], nums[curr]
                right -= 1
            else:
                curr += 1

        # 交叉合并
        small, big, _nums = mid, n - 1, nums[:]
        for i in range(n):
            if i % 2 == 0:
                nums[i] = _nums[small]
                small -= 1
            else:  # big
                nums[i] = _nums[big]
                big -= 1


# leetcode submit region end(Prohibit modification and deletion)

3.lc215topk

class Solution:
    def findKthLargest(self, nums: List[int], k: int) -> int:
        def findTopKth(low, high):
            pivot = random.randint(low, high)
            nums[low], nums[pivot] = nums[pivot], nums[low]
            base = nums[low]
            i = low
            j = low + 1
            while j <= high:
                if nums[j] > base:
                    nums[i + 1], nums[j] = nums[j], nums[i + 1]
                    i += 1
                j += 1
            nums[low], nums[i] = nums[i], nums[low]
            if i == k - 1:
                return nums[i]
            elif i > k - 1:
                return findTopKth(low, i - 1)
            else:
                return findTopKth(i + 1, high)
        return findTopKth(0, len(nums) - 1)