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.
代码
public class Solution {
public ListNode partition(ListNode node, int x) {
ListNode leftDummy = new ListNode(0);
ListNode leftCurr = leftDummy;
ListNode rightDummy = new ListNode(9);
ListNode rightCurr = rightDummy;
ListNode runner = node;
while (runner != null) {
if (runner.val < x) {
leftCurr.next = runner;
letCurr = leftCurr.next;
} else {
rightCurr.next = runner;
rithCurr = rightCurr.next;
}
runner = runner.next;
}
rightCurr.next = null;
leftCurr.next = rightDummy.next;
return leftDummy.next;
}
}