merge Sort

Author: yuzhangcmu

sort/MergeSort_LinkedList.java

@@ -0,0 +1,99 @@
+package Algorithms.sort;
+
+import java.util.LinkedList;
+
+import Algorithms.ListNode;
+
+public class MergeSort_LinkedList {
+    public static void main(String[] args) {
+        ListNode node1 = new ListNode(0);
+        ListNode node2 = new ListNode(15);
+        ListNode node3 = new ListNode(6);
+        ListNode node4 = new ListNode(9);
+        ListNode node5 = new ListNode(4);
+        
+        node1.next = node2;
+        node2.next = node3;
+        node3.next = node4;
+        node4.next = node5;
+        
+        System.out.println(node1.toString());
+        sort(node1);
+        System.out.println(node1.toString());
+    }
+    
+    public static ListNode sort(ListNode head) {
+        if (head == null) {
+            return null;
+        }
+        
+        // 注意一定要加这一行，否则会不断对1个元素无限分下去
+        if (head.next == null) {
+            return head;
+        }
+        
+        ListNode mid = findMidPre(head);
+        
+        // 将list切为2个list.
+        ListNode right = mid.next;
+        mid.next = null;
+        
+        //调用将2边分别排序
+        ListNode left = sort(head);
+        right = sort(right);
+        
+        // 将2个已经排序的List Merge在一起 
+        return merge(left, right); 
+    }
+    
+    public static ListNode merge(ListNode head1, ListNode head2) {
+        ListNode dummy = new ListNode(0);
+
+        // cur 表示新链的尾部.
+        ListNode cur = dummy;
+        
+        while (head1 != null && head2 != null) {
+            // 将2个链中较小的一个接到新链的尾部
+            if (head1.val < head2.val) {                
+                cur.next = head1;
+                head1 = head1.next;
+            } else {
+                cur.next = head2;
+                head2 = head2.next;
+            }
+
+            // 将扫描指针移动到新链的尾部0
+            cur = cur.next;
+        }        
+
+        // 把未扫描完毕的链接在新链的结尾即可
+        if (head1 != null) {
+            cur.next = head1;
+        } else {
+            cur.next = head2;
+        }
+        
+        // 返回新链的头部 
+        return dummy.next;   
+    }
+    
+    // 这个函数是寻找Mid的前一个节点，
+    // 技巧就是：一开始就将Fast放在head的前一个节点，这样当只有2个节点的时候：
+    // 1->2->null
+    // slow 会停在1处 ，这样就可以处理只有2个节点的情况.
+    public static ListNode findMidPre(ListNode head) {
+        if (head == null) {
+            return null;
+        }
+        
+        ListNode fast = head.next;
+        ListNode slow = head;
+        
+        if (fast != null && fast.next != null) {
+            fast = fast.next.next;
+            slow = slow.next;
+        }
+
+        return slow;
+    }
+}