86. Partition List Medium
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.
Example:
Input: head = 1->4->3->2->5->2, x = 3 Output: 1->2->2->4->3->5
梳理一下思路，应该很快可以写出来这样一个架子：
- 将来两个链表，分别存储分开之后的两个链表
- 遍历给定的链表，所有小于x的都放在前面的链表里；否则则加入到后面的链表
public ListNode partition(ListNode head, int x) {
ListNode before = new ListNode(0);
ListNode after = new ListNode(0);
ListNode bh = before, ah = after;
while (head != null) {
int v = head.val;
if(v < x) {
//TODO
}else {
//TODO
}
head = head.next;
}
return null;
}
到这里之后，如果操作链表就越写越糊涂了:(
链表操作：挂上节点；然后移动到链表尾部。
Runtime: 0 ms, faster than 100.00% of Java online submissions for Partition List. Memory Usage: 36 MB, less than 100.00% of Java online submissions for Partition List.
public ListNode partition(ListNode head, int x) {
ListNode before = new ListNode(0);
ListNode after = new ListNode(0);
ListNode bh = before, ah = after;
while (head != null) {
int v = head.val;
if(v < x) {
bh.next = head; //添加到前一个链表里
bh = bh.next; //移动到前一个链表的尾部
}else {
ah.next = head; //添加到后一个链表里
ah = ah.next; //移动到后一个链表的尾部
}
head = head.next;
}
ah.next = null; //确保断开后一个链表； 避免ah.next是一个小于x的节点时发生循环
bh.next = after.next; //连接两个链表：将前一个链表的尾部挂到后面的链表上
return before.next;
}