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leetcode--Partition List

2024-07-07 18:10:01   226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.

?
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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    /**This is a fundamental problem. But we need to pay attention to some subtle problems.
     * such as the null pointers.
     * @author Averill Zheng
     * @version 2016-06-07
     * @since JDK 1.7
     */
    public ListNode partition(ListNode head, int x) {
        ListNode newHead = new ListNode(0);
        if(head != null){
            newHead.next = head;
            ListNode firstLarger = head;
            ListNode insertPosition = newHead;
            while(firstLarger != null && firstLarger.val < x){
                insertPosition = insertPosition.next;
                firstLarger = firstLarger.next;
            }
            ListNode current = firstLarger;
            while(firstLarger != null){
                if(firstLarger.val >= x){
                    current = firstLarger;
                    firstLarger = firstLarger.next;
                }
                else{
                    current.next = firstLarger.next;
                    firstLarger.next = insertPosition.next;
                    insertPosition.next = firstLarger;
                    insertPosition = insertPosition.next;
                    firstLarger = current.next;
                }
            }
        }
        newHead = newHead.next;
        return newHead;
    }
}

　　


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