Leetcode - Partition List.

最新推荐文章于 2023-08-15 12:49:47 发布

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本文介绍了解决链表中根据给定值x进行分区的算法，将链表分为两部分，一部分包含所有小于x的节点，另一部分包含所有大于等于x的节点。文章详细解释了解题思路和代码实现。

https://oj.leetcode.com/problems/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.

public ListNode partition(ListNode head, int x)

这一题的解题思路很简单，就是根据x值构造两个list，一个list小于x，另一个list大于等于x即可，用一个tmp指针一路往下扫一路构建。然后连在一起就可以了。需要注意的不过是一些边界条件，譬如说x大于或者小于链表上的所有数。

    public ListNode partition(ListNode head, int x) {
        ListNode smallHead = null, smalltmp = null, largeHead = null, largetmp = null, tmp = head;
        while(tmp != null){
            if(tmp.val < x){
                if(smallHead == null){
                    smallHead = tmp;
                    smalltmp = smallHead;
                }else{
                    smalltmp.next = tmp;
                    smalltmp = smalltmp.next;
                }
            }else{
                if(largeHead == null){
                    largeHead = tmp;
                    largetmp = largeHead;
                }else{
                    largetmp.next = tmp;
                    largetmp =largetmp.next;
                }
            }
            tmp = tmp.next;
            if(smalltmp != null)smalltmp.next = null;
            if(largetmp != null)largetmp.next = null;
        }
        if(smalltmp != null)smalltmp.next = largeHead;
        return smallHead == null ? largeHead : smallHead;