归并排序:
public class Solution {
private ListNode findMiddle(ListNode head) {
ListNode slow = head, fast = head.next;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
private ListNode merge(ListNode head1, ListNode head2) {
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
while (head1 != null && head2 != null) {
if (head1.val < head2.val) {
tail.next = head1;
head1 = head1.next;
} else {
tail.next = head2;
head2 = head2.next;
}
tail = tail.next;
}
if (head1 != null) {
tail.next = head1;
} else {
tail.next = head2;
}
return dummy.next;
}
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode mid = findMiddle(head);
ListNode right = sortList(mid.next);
mid.next = null;
ListNode left = sortList(head);
return merge(left, right);
}
}
快排:
Version1:
public class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode mid = findMedian(head); // O(n)
ListNode leftDummy = new ListNode(0), leftTail = leftDummy;
ListNode rightDummy = new ListNode(0), rightTail = rightDummy;
ListNode middleDummy = new ListNode(0), middleTail = middleDummy;
while (head != null) {
if (head.val < mid.val) {
leftTail.next = head;
leftTail = head;
} else if (head.val > mid.val) {
rightTail.next = head;
rightTail = head;
} else {
middleTail.next = head;
middleTail = head;
}
head = head.next;
}
leftTail.next = null;
middleTail.next = null;
rightTail.next = null;
ListNode left = sortList(leftDummy.next);
ListNode right = sortList(rightDummy.next);
return concat(left, middleDummy.next, right);
}
private ListNode findMedian(ListNode head) {
ListNode slow = head, fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
private ListNode concat(ListNode left, ListNode middle, ListNode right) {
ListNode dummy = new ListNode(0), tail = dummy;
tail.next = left; tail = getTail(tail);
tail.next = middle; tail = getTail(tail);
tail.next = right; tail = getTail(tail);
return dummy.next;
}
private ListNode getTail(ListNode head) {
if (head == null) {
return null;
}
while (head.next != null) {
head = head.next;
}
return head;
Version2:
class Pair {
public ListNode first, second;
public Pair(ListNode first, ListNode second) {
this.first = first;
this.second = second;
}
}
public class Solution {
/**
* @param head: The head of linked list.
* @return: You should return the head of the sorted linked list,
using constant space complexity.
*/
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode mid = findMedian(head); // O(n)
Pair pair = partition(head, mid.val); // O(n)
ListNode left = sortList(pair.first);
ListNode right = sortList(pair.second);
getTail(left).next = right; // O(n)
return left;
}
// 1->2->3 return 2
// 1->2 return 1
private ListNode findMedian(ListNode head) {
ListNode slow = head, fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
// < value in the left, > value in the right
private Pair partition(ListNode head, int value) {
ListNode leftDummy = new ListNode(0), leftTail = leftDummy;
ListNode rightDummy = new ListNode(0), rightTail = rightDummy;
ListNode equalDummy = new ListNode(0), equalTail = equalDummy;
while (head != null) {
if (head.val < value) {
leftTail.next = head;
leftTail = head;
} else if (head.val > value) {
rightTail.next = head;
rightTail = head;
} else {
equalTail.next = head;
equalTail = head;
}
head = head.next;
}
leftTail.next = null;
rightTail.next = null;
equalTail.next = null;
if (leftDummy.next == null && rightDummy.next == null) {
ListNode mid = findMedian(equalDummy.next);
leftDummy.next = equalDummy.next;
rightDummy.next = mid.next;
mid.next = null;
} else if (leftDummy.next == null) {
leftTail.next = equalDummy.next;
} else {
rightTail.next = equalDummy.next;
}
return new Pair(leftDummy.next, rightDummy.next);
}
private ListNode getTail(ListNode head) {
if (head == null) {
return null;
}
while (head.next != null) {
head = head.next;
}
return head;
}
}
插入排序:
class Solution {
public:
/**
* @param head: The first node of linked list.
* @return: The head of linked list.
*/
ListNode *insertionSortList(ListNode *head) {
ListNode *dummy = new ListNode(0);
while (head != NULL) {
ListNode *temp = dummy;
ListNode *next = head->next;
while (temp->next != NULL && temp->next->val < head->val) {
temp = temp->next;
}
head->next = temp->next;
temp->next = head;
head = next;
}
return dummy->next;
}
};
本文详细解析了三种经典排序算法——归并排序、快速排序和插入排序的实现过程,并通过实例代码进行演示。通过对每种算法的时间复杂度、空间复杂度和稳定性进行对比分析,旨在为读者提供全面的理解,帮助选择最适合特定场景的排序方法。
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