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) { if (head == NULL) return NULL; ListNode lessHead(99), largeHead(99); ListNode* less = &lessHead, *large = &largeHead, *cur = head; while (cur != NULL) { if (cur->val < x) { less->next = cur; less = cur; } else { large->next = cur; large = cur; } cur = cur->next; } large->next = NULL; less->next = largeHead.next; return lessHead.next; } };