diff --git a/SortList.cpp b/SortList.cpp
index 01805b7..c540e7e 100644
--- a/SortList.cpp
+++ b/SortList.cpp
@@ -1,4 +1,13 @@
-#include "LeetCode.h"
+/*
+ Author:     cshony
+ Date:       May 19, 2015
+ Problem:    Sort List
+ Difficulty: Medium
+ Source:     https://leetcode.com/problems/sort-list/
+ Notes:
+ Sort a linked list in O(nlogn) time using constant space complexity.
+ Solution: merge sort without recursion.
+ */
 
 /**
  * Definition for singly-linked list.
@@ -7,28 +16,43 @@
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
- */
+ */ 
 class Solution {
 public:
-  ListNode *sortList(ListNode *head) {
-    if (head == NULL) return NULL;
-    ListNode h1(0), h2(0), h3(0), *t1=&h1, *t2=&h2, *t3=&h3;
-    int val = head->val;
-    while(head != NULL) {
-      if (head->val < val) {
-        t1 = t1->next = head;
-      } else if (head->val == val) {
-        t2 = t2->next = head;
-      } else {
-        t3 = t3->next = head;
-      }
-      head = head->next;
-    }
-    t1->next = t2->next = t3->next = NULL;
-    h1.next = sortList(h1.next);
-    for(t1=&h1; t1->next; t1=t1->next);
-    t1->next = h2.next;
-    t2->next = sortList(h3.next);
-    return h1.next;
-  }
+	ListNode* mergeList(ListNode *pre, ListNode *head1, ListNode *head2) {
+		while(head1 || head2) {
+			if(!head1) pre->next = head2, head2 = head2->next;
+			else if(!head2) pre->next = head1, head1 = head1->next;
+			else if(head1->val > head2->val) pre->next = head2, head2 = head2->next;
+			else pre->next = head1, head1 = head1->next;
+			pre = pre->next;
+		}
+		return pre;
+	}
+	ListNode* sortList(ListNode *head) {
+		ListNode fakeHead(0);
+		fakeHead.next = head;
+		int mergeLen = 1, mergeCount = 2;
+		while(mergeCount > 1) {
+			ListNode *cur = fakeHead.next, *pre = &fakeHead;
+			mergeCount = 0;
+			while(cur) {
+				mergeCount++;
+				if(!cur->next) { pre->next = cur; cur = NULL; continue; }
+				ListNode *leftEnd = cur, *rightEnd = cur->next;
+				int p = mergeLen;
+				while(--p && leftEnd->next) {
+					leftEnd = leftEnd->next;
+					rightEnd = rightEnd->next ? rightEnd->next : rightEnd;
+					rightEnd = rightEnd->next ? rightEnd->next : rightEnd;
+				}
+				ListNode *head1 = cur, *head2 = leftEnd->next;
+				cur = rightEnd->next;
+				leftEnd->next = rightEnd->next = NULL;
+				pre = mergeList(pre, head1, head2);
+			}
+			mergeLen*=2;
+		}
+		return fakeHead.next;
+	}
 };