[LeetCode] 86. Partition List

Given the head of 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.

Example 1:

Input: head = [1,4,3,2,5,2], x = 3
Output: [1,2,2,4,3,5]

Example 2:

Input: head = [2,1], x = 2
Output: [1,2]

Constraints:

• The number of nodes in the list is in the range [0, 200].
• -100 <= Node.val <= 100
• -200 <= x <= 200

分隔链表。

给你一个链表和一个特定值 x ，请你对链表进行分隔，使得所有小于 x 的节点都出现在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/partition-list
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思路是设两个node，smallHead, bigHead，然后开始遍历链表，大于X和小于X的 node 分别加到 smallHead 和 bigHead 后面。遍历完链表之后将两个链表再连起来。注意最后需要把 bigHead 的 next 指针设置为 NULL，因为当前节点复用的是原链表的节点，而其 next 指针可能指向一个小于X的节点。

时间O(n)

空间O(1)

JavaScript实现

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} x
 * @return {ListNode}
 */
var partition = function (head, x) {
    // corner case
    if (head === null || head.next === null) {
        return head;
    }

    // normal case
    let pHead = new ListNode(0);
    let p = pHead;
    let qHead = new ListNode(0);
    let q = qHead;
    while (head !== null) {
        if (head.val < x) {
            p.next = head;
            p = p.next;
        } else {
            q.next = head;
            q = q.next;
        }
        head = head.next;
    }
    p.next = qHead.next;
    q.next = null;
    return pHead.next;
};

Java实现

class Solution {
    public ListNode partition(ListNode head, int x) {
        // corner case
        if (head == null || head.next == null) {
            return head;
        }

        // normal case
        ListNode p = new ListNode(0);
        ListNode pHead = p;
        ListNode q = new ListNode(0);
        ListNode qHead = q;
        while (head != null) {
            // < x成一组
            if (head.val < x) {
                p.next = head;
                p = p.next;
            }
            // >= x成一组
            else {
                q.next = head;
                q = q.next;
            }
            head = head.next;
        }
        p.next = qHead.next;
        q.next = null;
        return pHead.next;
    }
}

LeetCode 题目总结