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LeetCode: Partition List
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.
地址：https://oj.leetcode.com/problems/partition-list/
算法：首先，找到第一个大于等于x的节点，记其前趋节点为pre，则在继续往后遍历，若发现比x小的值，则把该节点从链表中移出来，插入到pre节点后面。代码：
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) return NULL;13 ListNode *p = head;14 ListNode *pre = NULL;15 while(p && p->val < x){16 pre = p;17 p = p->next;18 }19 if(!p)20 return head;21 ListNode *q = NULL;22 while(p->next){23 if(p->next->val < x){24 q = p->next;25 p->next = q->next;26 if(pre){27 q->next = pre->next;28 pre->next = q;29 pre = q;30 }else{31 q->next = head;32 head = q;33 pre = q;34 }35 }else{36 p = p->next;37 }38 }39 return head;40 }41 };
LeetCode: Partition List
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