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86. Partition List

2024-09-10 23:02:23   215人阅读

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.

该题将链表拆成两个，left里存小于x的结点，right里存大于等于x的结点，最后将left和right拼接起来

public class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head == null) {
            return null;
        }
        
        ListNode leftDummy = new ListNode(0);
        ListNode rightDummy = new ListNode(0);
        ListNode left = leftDummy, right = rightDummy;
        //将小于x的结点加入到left里，大于等于x的结点加入到right里
        while (head != null) {
            if (head.val < x) {
                left.next = head;
                left = head;
            } else {
                right.next = head;
                right = head;
            }
            head = head.next;
        }
        //记得给right.next赋值为null
        right.next = null;
        left.next = rightDummy.next;
        return leftDummy.next;
    }
}

 

86. Partition List


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