Leetcode: Partition List
思路: 1. 空间复杂度为 o(n) 解法. 创建两个链表, 分别记录大于 x 和小于 x 的节点, 最后合并 2. o(1) 的空间复杂度解法. 四个指针, 分别指向小于 x 部分链表的头, 尾, 指向大于 x 部分链表的头, 尾 总结: 1. 使用 dummyNode 减少判断 ...
原文：[LeetCode] Partition List 划分链表
Given a linked list and a valuex, partition it such that all nodes less thanxcome before nodes greater than or equal tox. You should preserve the original relative order of the nodes in each of the t ...
2015-03-08 08:37 0 11175 推荐指数：
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