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Partition List
题目
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,
Given1->4->3->2->5->2
and x = 3,
return1->2->2->4->3->5
.
方法
将原来的Linked List 拆分成两个, 一个记录比X小的元素,另一个记录比X大的元素,最后将其合并。public ListNode partition(ListNode head, int x) { ListNode lessStart = null; ListNode lessEnd = null; ListNode greaterStart = null; ListNode greaterEnd = null; ListNode cur = head; while (cur != null) { if (cur.val < x) { if (lessStart == null) { lessStart = cur; lessEnd = cur; } else { lessEnd.next = cur; lessEnd = cur; } } else { if (greaterStart == null) { greaterStart = cur; greaterEnd = cur; } else { greaterEnd.next = cur; greaterEnd = cur; } } cur = cur.next; } if (lessStart == null) { return greaterStart; } if (greaterStart == null) { return lessStart; } greaterEnd.next = null; lessEnd.next = greaterStart; return lessStart; }
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