给出一个链表和一个值x,以x为参照将链表划分成两部分,使所有小于x的节点都位于大于或等于x的节点之前。
两个部分之内的节点之间要保持的原始相对顺序。
例如:
给出1->4->3->2->5->2和x = 3,
返回1->2->2->4->3->5
思路:这个题目比较简单,搞两个链表分别存储两部分,然后链接起来即可
<?php
class Node {
public $next = null;
public $val;
public function __construct($val) {
$this->val = $val;
}
}
function partition($head, $x) {
$smallNode = new Node(0);
$largeNode = new Node(0);
$largeNode->next = $head;
$node = $smallNode;
$pre = $largeNode;
while ($head != null) {
if ($head->val < $x) {
$tmp = $head->next;
$head->next = $node->next;
$node->next = $head;
$pre->next = $tmp;
$node = $node->next;
$head = $tmp;
} else {
$pre = $head;
$head = $head->next;
}
}
$node->next = $largeNode->next;
return $smallNode->next;
}
$node1 = new Node(1);
$node2 = new Node(4);
$node3 = new Node(3);
$node4 = new Node(2);
$node5 = new Node(5);
$node6 = new Node(2);
$node1->next = $node2;
$node2->next = $node3;
$node3->next = $node4;
$node4->next = $node5;
$node5->next = $node6;
$retNode = partition($node1, 3);
while ($retNode != null) {
print $retNode->val . "\n";
$retNode = $retNode->next;
}