refactor: Replace 结点 with 节点 (#452)
* Replace 结点 with 节点 Update the footnotes in the figures * Update mindmap * Reduce the size of the mindmap.png
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桶排序的时间复杂度理论上可以达到 $O(n)$ ,**难点是需要将元素均匀分配到各个桶中**,因为现实中的数据往往都不是均匀分布的。举个例子,假设我们想要把淘宝的所有商品根据价格范围平均分配到 10 个桶中,然而商品价格不是均匀分布的,100 元以下非常多、1000 元以上非常少;如果我们将价格区间平均划为 10 份,那么各个桶内的商品数量差距会非常大。
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为了实现平均分配,我们可以先大致设置一个分界线,将数据粗略分到 3 个桶,分配完后,**再把商品较多的桶继续划分为 3 个桶,直至所有桶内元素数量大致平均为止**。此方法本质上是生成一个递归树,让叶结点的值尽量平均。当然,不一定非要划分为 3 个桶,可以根据数据特点灵活选取。
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为了实现平均分配,我们可以先大致设置一个分界线,将数据粗略分到 3 个桶,分配完后,**再把商品较多的桶继续划分为 3 个桶,直至所有桶内元素数量大致平均为止**。此方法本质上是生成一个递归树,让叶节点的值尽量平均。当然,不一定非要划分为 3 个桶,可以根据数据特点灵活选取。
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观察发现,归并排序的递归顺序就是二叉树的「后序遍历」。
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- **后序遍历**:先递归左子树、再递归右子树、最后处理根结点。
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- **后序遍历**:先递归左子树、再递归右子树、最后处理根节点。
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- **归并排序**:先递归左子树、再递归右子树、最后处理合并。
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=== "Java"
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归并排序有一个很特别的优势,用于排序链表时有很好的性能表现,**空间复杂度可被优化至 $O(1)$** ,这是因为:
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- 由于链表可仅通过改变指针来实现结点增删,因此“将两个短有序链表合并为一个长有序链表”无需使用额外空间,即回溯合并阶段不用像排序数组一样建立辅助数组 `tmp` ;
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- 由于链表可仅通过改变指针来实现节点增删,因此“将两个短有序链表合并为一个长有序链表”无需使用额外空间,即回溯合并阶段不用像排序数组一样建立辅助数组 `tmp` ;
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- 通过使用「迭代」代替「递归划分」,可省去递归使用的栈帧空间;
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> 详情参考:[148. 排序链表](https://leetcode-cn.com/problems/sort-list/solution/sort-list-gui-bing-pai-xu-lian-biao-by-jyd/)
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