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) { // Start typing your C/C++ solution below // DO NOT write int main() function if(head == NULL) return head; ListNode *pre , *firstLarger, * preCur, * cur; pre = NULL; firstLarger = head; while(firstLarger && firstLarger->val < x){ pre = firstLarger; firstLarger = firstLarger ->next; } if(firstLarger == NULL) return head; preCur = firstLarger; cur = preCur->next; while(cur != NULL){ if( cur->val >= x){ preCur = cur; cur = cur->next; continue; } preCur->next = cur->next; cur->next = firstLarger; if(pre == NULL){ pre = cur; head = cur; }else{ pre->next = cur; pre = cur; } cur = preCur ->next; } return head; } };