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[LeetCode] Partition List
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.
思路：新建两个头结点，分别把小于x 的节点链接到第一个头结点后，把大于或等于x 的节点链接到第二个头节点后。
时间复杂度O(n),空间复杂度O(1)
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution {10 public:11 ListNode *partition(ListNode *head, int x) {12 if(head == NULL) return head;13 14 ListNode *plow = new ListNode(-1);15 ListNode *plhead = plow;16 ListNode *phigh = new ListNode(-1);17 ListNode *phhead = phigh;18 ListNode *pt = head;19 20 while (pt != NULL) {21 if (pt->val < x) {22 plow->next = pt;23 plow = pt;24 } else {25 phigh->next = pt;26 phigh = pt;27 }28 pt = pt->next;29 }30 31 plow->next = phhead->next;32 phigh->next = NULL;33 34 return plhead->next;35 }36 };
[LeetCode] Partition List
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