[LeetCode]Partition List, 解题报告

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,

Given 1->4->3->2->5->2 and x = 3,

return 1->2->2->4->3->5.

思路

题意：这道题目就是在不改变链表原来顺序的情况下，将大于x的节点移到后面，小于x的节点移到前面

解法：我们可以构建两个链表，一个链表存放所有小于x的节点，另一个链表存放所有大于x的节点，然后连接两个链表即可

AC代码

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode partition(ListNode head, int x) {
ListNode smaller, bigger, tmps, tmpb;

smaller = new ListNode(0);
tmps = smaller;

bigger = new ListNode(0);
tmpb = bigger;

while (head != null) {
if (head.val < x) {
tmps.next = head;
tmps = head;
} else {
tmpb.next = head;
tmpb = head;
}

head = head.next;
}

tmpb.next = null;
tmps.next = bigger.next;

return smaller.next;
}
}