分隔链表
题目介绍
给你一个链表的头节点 head 和一个特定值 x ，请你对链表进行分隔，使得所有 小于 x 的节点都出现在 大于或等于 x 的节点之前。
你应当 保留 两个分区中每个节点的初始相对位置。
示例1：
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| 输入：head = [1,4,3,2,5,2], x = 3
输出：[1,2,2,4,3,5]
|
示例 2：
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| 输入：head = [2,1], x = 2
输出：[1,2]
|
提示：
- 链表中节点的数目在范围
[0, 200] 内
-100 <= Node.val <= 100-200 <= x <= 200
题目解法
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| package algorithm;
public class PartitionList {
public static ListNode partition(ListNode head, int x) {
ListNode small = new ListNode(0);
ListNode smallHead = small;
ListNode large = new ListNode(0);
ListNode largeHead = large;
while (head != null) {
if (head.val < x) {
small.next = head;
small = small.next;
} else {
large.next = head;
large = large.next;
}
head = head.next;
}
large.next = null;
small.next = largeHead.next;
return smallHead.next;
}
public static void main(String[] args) {
ListNode node1 = new ListNode(1);
ListNode node2 = new ListNode(4);
node1.next = node2;
ListNode node3 = new ListNode(3);
node2.next = node3;
ListNode node4 = new ListNode(2);
node3.next = node4;
ListNode node5 = new ListNode(5);
node4.next = node5;
ListNode node6 = new ListNode(2);
node5.next = node6;
node6.next = null;
print(partition(node1, 3));
node1 = new ListNode(2);
node2 = new ListNode(1);
node1.next = node2;
node2.next = null;
print(partition(node1, 2));
}
private static void print(ListNode head) {
while (head != null) {
System.out.print(head.val + " ");
head = head.next;
}
System.out.println();
}
public static class ListNode {
int val;
ListNode next;
ListNode() {
}
ListNode(int val) {
this.val = val;
}
ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}
}
|
打印：
思路：
思路上，双链表实在太巧妙了。