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[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.
/** * 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 *p = head; ListNode *lessHead = NULL,*greaterHead =NULL,*pLess=NULL,*pGreater=NULL; while(p){ if(p->val < x ){ if(lessHead == NULL){ lessHead = p; pLess = p; }else{ pLess->next = p; pLess = pLess->next; } }else{ if(greaterHead == NULL){ greaterHead = p ; pGreater = p; }else{ pGreater->next = p; pGreater = pGreater->next; } } p = p->next; }//end while if(pGreater!=NULL) pGreater->next = NULL; if(pLess !=NULL){ pLess->next = greaterHead; return lessHead; }else{ return greaterHead; } }//end func };
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