148. Sort List | 易学教程

题目：

Sort a linked list in O(n log n) time using constant space complexity.

链接： http://leetcode.com/problems/sort-list/

题解：

Sort List，链表排序，一看到O(nlogn)就想到使用merge sort。对于merge sort，还要研究一下bottom-up和top-down的区别，优劣点，如何继续优化等等，比如n较小时使用insertion sort，了解Python的Tim-sort之类。也要研究一下链表的Quicksort，以及3-way Quicksort。当时电面Amazon的时候写排序作死选择Quicksort，还尝试写了一个3-way Quicksort，由于理解不深同时当天也头昏脑胀没解释清楚，被面试官白大姐秒拒，如此浪费一年机会，甚是可惜。对于各种排序的stalability也要好好复习。熟能生巧，多练吧。另外，external sorting，B tree，A* tree，indexing，paging之类的也要复习。聊远了...

这题使用Merge Sort的话，还是按照Divide-and-Conquer分治。两个辅助方法是findMid找中点，以及merge合并。 merge的话完全可以使用"Merge Sorted List"的代码，找中点我们也使用过，所以可以直接写。

Time Complexity - O(nlogn)， Space Complexity - O(logn)。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null || head.next == null)
            return head;
        ListNode mid = findMid(head);
        ListNode right = sortList(mid.next);
        mid.next = null;
        ListNode left = sortList(head);
        
        
        return merge(left, right);
    }
    
    private ListNode findMid(ListNode head) {　　　　　　　　
        if(head == null || head.next == null)
            return head;
        ListNode fast = head, slow = head;
        
        while(fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        
        return slow;
    }
    
    private ListNode merge(ListNode left, ListNode right) {
        ListNode dummy = new ListNode(-1);
        ListNode node = dummy;
        
        while(left != null && right != null) {
            if(left.val < right.val) {
                node.next = left;
                left = left.next;
            } else {
                node.next = right;
                right = right.next;
            }
            node = node.next;
        }
        
        if(left != null)
            node.next = left;
        else
            node.next = right;
            
        return dummy.next;
    }
}

到这里其实还没结束。仔细看题目要求constant space complexity。这样的话我们就不能使用递归了。

二刷:

O(1) space并不好写，此刷没有完成，也并没有尝试quicksort，很忧伤... 最后还是使用的merge sort recrusive的方法。代码大都参考了Discuss区里zwangbo同学的。

Java:

Time Complexity - O(nlogn)， Space Complexity - O(logn)。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) return head; 
        ListNode fast = head, slow = head, preMid = head;
        while (fast != null && fast.next != null) {
            preMid = slow;
            fast = fast.next.next;
            slow = slow.next;
        }
        preMid.next = null;
        return merge(sortList(head), sortList(slow));
    }
    
    private ListNode merge(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        if (l1.val < l2.val) {
            l1.next = merge(l1.next, l2);
            return l1;
        } else {
            l2.next = merge(l1, l2.next);
            return l2;
        }
    }
}

三刷:

依然是使用之前的方法，找中点，再递归进行mergesort。

Java:

Time Complexity - O(nlogn)， Space Complexity - O(logn)。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) return head;
        ListNode fast = head, slow = head, prev = head;
        
        while (fast != null && fast.next != null) {
            prev = slow;
            fast = fast.next.next;
            slow = slow.next;
        }
        prev.next = null;
        return mergeTwoSortedList(sortList(head), sortList(slow));
    }
    
    private ListNode mergeTwoSortedList(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode dummy = new ListNode(-1);
        ListNode curNode = dummy, p1 = l1, p2 = l2;
        
        while (p1 != null && p2 != null) {
            if (p1.val < p2.val) {
                curNode.next = p1;
                p1 = p1.next;
            } else {
                curNode.next = p2;
                p2 = p2.next;
            }
            curNode = curNode.next;
        }
        curNode.next = (p1 != null) ? p1 : p2;
        return dummy.next;
    } 
    
}

Reference：

http://www.mathcs.emory.edu/~cheung/Courses/171/Syllabus/7-Sort/merge-sort5.html

https://en.wikipedia.org/wiki/Merge_sort#Bottom-up_implementation

https://leetcode.com/discuss/10264/my-o-n-log-n-time-o-1-space-solution

https://leetcode.com/discuss/12014/my-iterative-merge-sort

https://leetcode.com/discuss/15420/java-solution-with-strict-o-1-auxiliary-space-complexity

https://leetcode.com/discuss/18210/i-think-my-java-solution-is-o-1-space

https://leetcode.com/discuss/19119/java-merge-sort-solution-with-o-1-memory

https://leetcode.com/discuss/24917/my-java-code-with-constant-space

https://leetcode.com/discuss/28594/bottom-recurring-with-space-complextity-time-complextity

https://leetcode.com/discuss/39867/java-solution-bottom-iterative-merge-with-detailed-comments

https://leetcode.com/discuss/1709/have-pretty-mergesort-method-anyone-speed-reduce-memory-usage

https://leetcode.com/discuss/92701/java-solution-using-merge-sort

https://leetcode.com/discuss/44369/java-merge-sort-solution

https://leetcode.com/discuss/29921/basically-seems-merge-sort-problem-really-easy-understand

https://leetcode.com/discuss/81455/java-quick-sort-fast-beats-98%25-also-includes-merge-sort-code

https://leetcode.com/discuss/81263/super-easy-understand-java-iterative-merge-sort-using-space

来源：https://www.cnblogs.com/yrbbest/p/4489582.html

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