标签:拓展 ext return 关于 lin 主函数 code this tin
一.通过X将链表排序;小的在前大的在后:class ListNode {
public int val;
public ListNode next;
public ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
public ListNode(int val) {
this(val, null);
}
}
public class LinkedListInterview {
public ListNode separateByX(ListNode head, int x) {
ListNode sHead = null;
ListNode sEnd = null;
ListNode bHead = null;
ListNode bEnd = null;
for (ListNode cur = head; cur != null; cur = cur.next) {
if (cur.val < x) {
if (sHead == null) {
sHead = cur;
} else {
sEnd.next = cur;
}
sEnd = cur;
} else {//≥
if (bHead == null) {
bHead = cur;
} else {
bEnd.next = cur;
}
bEnd = cur;
}
}
if (sEnd == null) {
return bHead;
}
sEnd.next = bHead;
if (bEnd != null) {
bEnd.next = null;
}
return sHead;
}
private static ListNode createTestList() {
ListNode n1 = new ListNode(4);
ListNode n2 = new ListNode(5);
ListNode n3 = new ListNode(2);
ListNode n4 = new ListNode(7);
ListNode n5 = new ListNode(6);
ListNode n6 = new ListNode(3);
ListNode n7 = new ListNode(8);
ListNode n8 = new ListNode(1);
n1.next = n2;
n2.next = n3;
n3.next = n4;
n4.next = n5;
n5.next = n6;
n6.next = n7;
n7.next = n8;
return n1;
}
// 三个重要节点
// 1. 咱创建的测试链表是不是出问题了?
// 2. 咱的分离程序是不是出问题了?
// 3. 咱的打印程序是不是出问题了?
private static void test() {
// 4 5 2 7 6 3 8 1
ListNode head = createTestList();
ListNode result = new LinkedListInterview().separateByX(head, 5);
// 4 2 3 1 5 7 6 8
for (ListNode cur = result; cur != null; cur = cur.next) {
System.out.println(cur.val);
}
}
public static void main(String[] args) {//主函数,调用test子函数
test();
}
}
标签:拓展 ext return 关于 lin 主函数 code this tin
原文地址:https://blog.51cto.com/14232658/2443903