leetcode 086 Partition List

最新推荐文章于 2019-07-16 09:33:43 发布

walter1990

最新推荐文章于 2019-07-16 09:33:43 发布

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本文介绍了一种链表分区算法，该算法将链表中小于给定值x的所有节点置于大于等于x的节点之前，同时保持各分区内的相对顺序不变。例如，对于链表1->4->3->2->5->2 和 x = 3，算法返回 1->2->2->4->3->5。通过遍历链表，并创建两个新的链表分别存储小于x和大于等于x的节点，最终将两链表合并。

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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.

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class Solution {
public:
    ListNode* partition(ListNode* head, int x) {
        ListNode *small=NULL, *greater=NULL, *smallCur=NULL, *greaterCur=NULL;
		
		if(head==NULL) return head;

		for(ListNode *cur=head; cur!=NULL;) {
		    ListNode *next = cur->next;
			if(cur->val < x) {
				if(small==NULL) {
					small=cur;
				} else {
					smallCur->next=cur;
				}
				smallCur=cur;
				smallCur->next = NULL;
			} else {
				if(greater==NULL) {
					greater=cur;
				} else {
					greaterCur->next=cur;
				}
				greaterCur=cur;
				greaterCur->next=NULL;
			}
			cur = next;
		}

		if(small!=NULL) {
			smallCur->next=greater;
		} else {
			small=greater;
		}

		return small;
    }
};