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
.
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *partition(ListNode *head, int x) { ListNode *newnode1 = new ListNode(-1); ListNode *newnode2 = new ListNode(-1); ListNode *smallr = newnode1; ListNode *large = newnode2; ListNode *cur = head; while(cur !=NULL){ if(cur->valnext= cur; smallr=smallr->next; }else{ large->next =cur; large=large->next; } cur=cur->next; } large->next=NULL; //處理好尾部節點 防止有環 smallr->next= newnode2->next; return newnode1->next; } };