LeetCode题解-86-Partition List

原题

解法1（参考别人，更简洁）

解题思路

将比X小的节点串成一串head1，将剩余的节点串成一串head2，最后将head1->head2

代码

public class Solution86 {public ListNode partition(ListNode head, int x) {if (head == null || head.next == null)return head;ListNode lessHead = new ListNode(0);ListNode moreOrEqualHead = new ListNode(0);ListNode iterator = head, lessIterator = lessHead, moreOrEqualIterator = moreOrEqualHead;while (iterator != null){if (iterator.val < x){lessIterator.next = iterator;lessIterator = lessIterator.next;}else {moreOrEqualIterator.next = iterator;moreOrEqualIterator = moreOrEqualIterator.next;}iterator = iterator.next;}moreOrEqualIterator.next = null;lessIterator.next = moreOrEqualHead.next;return lessHead.next;}}

解法2

解题思路

类似冒泡排序，将比x的值小的节点从尾部“冒泡”到前方：每一次将第i个节点插入到第0个到第i个节点之间的合适的位置，插入完毕后，前i个节点符合题目要求，执行N次，结束。

图解

Previous指向的为节点应该插入的位置，current节点与next节点用来判断节点是否需要移动以及存放移动时的信息。

代码

public class Solution {public ListNode partition(ListNode head, int x) {if (head == null || head.next == null)return head;ListNode dummy = new ListNode(0);dummy.next = head;ListNode previous = dummy, current;while(previous.next != null && previous.next.val < x){previous = previous.next;}current = previous;while (current.next != null){ListNode next = current.next;if (current.next.val < x && current.val >= x){current.next = next.next;next.next = previous.next;previous.next = next;previous = previous.next;}else {current = current.next;}}return dummy.next;}}