LeetCode 148. 排序链表
题目描述
思路分析
- 将链表分成两半。
- 对每一半递归地进行排序。
- 将两个排序好的链表合并成一个有序的链表。
参考代码
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func sortList(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
slow, fast := head, head
preSlow := &ListNode{Next: head}
for fast != nil && fast.Next != nil {
fast = fast.Next.Next
preSlow = slow
slow = slow.Next
}
preSlow.Next = nil
left := sortList(head)
right := sortList(slow)
return mergeTwoLists(left, right)
}
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
if l1 == nil {
return l2
} else if l2 == nil {
return l1
}
if l1.Val < l2.Val {
l1.Next = mergeTwoLists(l1.Next, l2)
return l1
} else {
l2.Next = mergeTwoLists(l1, l2.Next)
return l2
}
}
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func sortList(head *ListNode) *ListNode {
if head == nil || head.Next == nil {
return head
}
slow, fast := head, head
var pre *ListNode
for fast != nil && fast.Next != nil {
pre = slow
slow = slow.Next
fast = fast.Next.Next
}
pre.Next = nil
l1 := sortList(head)
l2 := sortList(slow)
return mergeTwoLists(l1, l2)
}
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
if l1 == nil {
return l2
} else if l2 == nil {
return l1
}
dummy := &ListNode{}
cur := dummy
for l1 != nil && l2 != nil {
if l1.Val < l2.Val {
cur.Next = l1
l1 = l1.Next
} else {
cur.Next = l2
l2 = l2.Next
}
cur = cur.Next
}
if l1 != nil {
cur.Next = l1
} else {
cur.Next = l2
}
return dummy.Next
}
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public class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode preSlow = head, slow = head, fast = head;
while (fast != null && fast.next != null) {
preSlow = slow;
slow = slow.next;
fast = fast.next.next;
}
preSlow.next = null;
ListNode l1 = sortList(head);
ListNode l2 = sortList(slow);
return mergeTwoLists(l1, l2);
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null || l2 == null) {
return l1 == null ? l2 : l1;
}
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
}
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