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Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
You should preserve the original relative order of the nodes in each of the two partitions.
For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.
这道题我本想用交换节点的方法处理，因为势必要保存当前节点的前一个节点，所以交换起来非常复杂
后来上网一查，发现可以直接使用创建两个链表然后合并的方法，不禁恍然大悟，忽略了链表的灵活性
既然partition方法可以很容易的实现，那么链表的快速排序也可以实现了，有时间再实现
ListNode* partition(ListNode* head, int x) { if (head == nullptr) return head; ListNode temp1(0); ListNode temp2(0); ListNode *cur1 = &temp1; ListNode *cur2 = &temp2; temp1.next = head; int tag = 0; while (head != nullptr) { if (head->val < x) { if (tag == 1) { cur1->next = head; tag = 0; } cur1 = cur1->next; } else if (head->val >= x) { if (tag == 0) { cur2->next = head; tag = 1; } cur2 = cur2->next; } head = head->next; } cur1->next = cur2->next = nullptr; cur1->next = temp2.next; return temp1.next; }
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原文地址：http://www.cnblogs.com/sdlwlxf/p/5073907.html
