148. Sort List

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

這道題是Linked List的排序，但是對空間複雜度是有要求的。
我首先把昨天學會的那個插入排序寫上了，本來以爲不會通過，沒想到竟然通過了。但是這個用時竟然有603ms，很嚇人。

class Solution {
    public ListNode sortList(ListNode head) {
        if(head == null) return head;
		ListNode helper = new ListNode(0);
		ListNode cur = head;
		ListNode pre = helper;
		ListNode next = null;
		while(cur != null) {
			next = cur.next;
			while(pre.next != null && pre.next.val < cur.val) {
				pre = pre.next;
			}
			cur.next = pre.next;
			pre.next = cur;
			pre = helper;
			cur = next;
		}
		return helper.next;
    }
}

然後我看了discuss，發現了點贊最多的一個回答。這個方法，用了分而治之和遞歸的思想，很多人說由於用了遞歸不滿足空間複雜度的要求。

class Solution {
    public ListNode sortList(ListNode head) {
	    if (head == null || head.next == null)
	      return head;
	        
	    // step 1. cut the list to two halves
	    ListNode prev = null, slow = head, fast = head;
	    
	    while (fast != null && fast.next != null) {
	      prev = slow;
	      slow = slow.next;
	      fast = fast.next.next;
	    }
	    
	    prev.next = null;
	    
	    // step 2. sort each half
	    ListNode l1 = sortList(head);
	    ListNode l2 = sortList(slow);
	    
	    // step 3. merge l1 and l2
	    return merge(l1, l2);
	  }
	  
	  ListNode merge(ListNode l1, ListNode l2) {
	    ListNode l = new ListNode(0), p = l;
	    
	    while (l1 != null && l2 != null) {
	      if (l1.val < l2.val) {
	        p.next = l1;
	        l1 = l1.next;
	      } else {
	        p.next = l2;
	        l2 = l2.next;
	      }
	      p = p.next;
	    }
	    
	    if (l1 != null)
	      p.next = l1;
	    
	    if (l2 != null)
	      p.next = l2;
	    
	    return l.next;
	  }

}

還看到了一種解法，這個是空間複雜度爲O(1)的。

class Solution {
    public ListNode sortList(ListNode head) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode [] list = new ListNode[2];
    boolean done = (null == head);
    // Keep partitioning our list into bigger sublists length. Starting with a size of 1 and doubling each time
    for (int step = 1; !done; step *= 2) {
      done = true;
      ListNode prev = dummy;
      ListNode remaining = prev.next;
      do {
        // Split off two sublists of size step
        for (int i = 0; i < 2; ++i) {
          list[i] = remaining;
          ListNode tail = null;
          for (int j = 0; j < step && null != remaining; ++j, remaining = remaining.next) {
            tail = remaining;
          }
          // Terminate our sublist
          if (null != tail) {
            tail.next = null;
          }
        }

        // We're done if these are the first two sublists in this pass and they
        // encompass the entire primary list
        done &= (null == remaining);

        // If we have two sublists, merge them into one
        if (null != list[1]) {
          while (null != list[0] || null != list[1]) {
            int idx = (null == list[1] || null != list[0] && list[0].val <= list[1].val) ? 0 : 1;
            prev.next = list[idx];
            list[idx] = list[idx].next;
            prev = prev.next;
          }

          // Terminate our new sublist
          prev.next = null;
        } else {
          // Only a single sublist, no need to merge, just attach to the end
          prev.next = list[0];
        }
      } while (null != remaining);
    }
    return dummy.next;
  }
}