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.
Solution：two pointers
思路：创建两个新list，将原list根据和x相比的大小剥离出来分别放到这两个list中，再连接
Time Complexity: O(N) Space Complexity: O(1)
Solution Code：
class Solution {
public ListNode partition(ListNode head, int x) {
ListNode dummy1 = new ListNode(0), dummy2 = new ListNode(0);
ListNode curr1 = dummy1, curr2 = dummy2;
while (head != null){
if (head.val < x) {
curr1.next = head;
curr1 = head;
}else {
curr2.next = head;
curr2 = head;
}
head = head.next;
}
curr2.next = null;
curr1.next = dummy2.next;
return dummy1.next;
}
}
