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,
Given1->4->3->2->5->2 and x = 3,return 1->2->2->4->3->5. Solution:
/** * 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) { if(head == NULL) return head; ListNode *ans_head = NULL,*ans_end = NULL, *tmp = head; while(tmp != NULL) { if(tmp -> val < x) { ListNode *ln = new ListNode(tmp -> val); if(ans_head == NULL) { ans_head = ln; ans_end = ln; } else { ans_end -> next = ln; ans_end = ln; } } tmp = tmp -> next; } tmp = head; while(tmp != NULL) { if(tmp -> val >= x) { ListNode *ln = new ListNode(tmp -> val); if(ans_head == NULL) { ans_head = ln; ans_end = ln; } else { ans_end -> next = ln; ans_end = ln; } } tmp = tmp -> next; } return ans_head; }}; 