solve problem Insertion Sort List

Author: P-ppc

README.md

@@ -253,6 +253,7 @@ All solutions will be accepted!
 |143|[Reorder List](https://leetcode-cn.com/problems/reorder-list/description/)|[java/py/js](./algorithms/ReorderList)|Medium|
 |515|[Find Largest Value In Each Tree Row](https://leetcode-cn.com/problems/find-largest-value-in-each-tree-row/description/)|[java/py/js](./algorithms/FindLargestValueInEachTreeRow)|Medium|
 |199|[Binary Tree Right Side View](https://leetcode-cn.com/problems/binary-tree-right-side-view/description/)|[java/py/js](./algorithms/BinaryTreeRightSideView)|Medium|
+|147|[Insertion Sort List](https://leetcode-cn.com/problems/insertion-sort-list/description/)|[java/py/js](./algorithms/InsertionSortList)|Medium|
 
 # Database
 |#|Title|Solution|Difficulty|

algorithms/InsertionSortList/README.md

@@ -0,0 +1,2 @@
+# Insertion Sort List
+We can solve this problem by dummy node

algorithms/InsertionSortList/Solution.java

@@ -0,0 +1,40 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode insertionSortList(ListNode head) {
+        ListNode dummy = new ListNode(-1);
+        dummy.next = head;
+        ListNode cur = dummy.next,
+                 pre = cur;
+        
+        while (cur != null) {
+            if (cur.val < pre.val) {
+                ListNode next = cur.next;
+                pre.next = next;
+                
+                ListNode p1 = dummy,
+                         p2 = p1.next;
+                
+                while (p2.val < cur.val) {
+                    p1 = p1.next;
+                    p2 = p2.next;
+                }
+                cur.next = p2;
+                p1.next = cur;
+                
+                cur = next;
+            } else {
+                pre = cur;
+                cur = cur.next;
+            }
+        }
+        
+        return dummy.next;
+    }
+}

algorithms/InsertionSortList/solution.js

@@ -0,0 +1,42 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var insertionSortList = function(head) {
+    let dummy = new ListNode(-1)
+    dummy.next = head
+    
+    let cur = dummy.next,
+        pre = cur
+    
+    while (cur) {
+        if (cur.val < pre.val) {
+            let next = cur.next
+            pre.next = next
+            
+            let p1 = dummy,
+                p2 = dummy.next
+            
+            while (p2.val < cur.val) {
+                p1 = p1.next
+                p2 = p2.next
+            }
+            cur.next = p2
+            p1.next = cur
+            
+            cur = next
+        } else {
+            pre = cur
+            cur = cur.next
+        }
+    }
+    
+    return dummy.next
+};

algorithms/InsertionSortList/solution.py

@@ -0,0 +1,39 @@
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution(object):
+    def insertionSortList(self, head):
+        """
+        :type head: ListNode
+        :rtype: ListNode
+        """
+        dummy = ListNode(-1)
+        dummy.next = head
+        
+        cur = dummy.next
+        pre = cur
+        
+        while cur:
+            if cur.val < pre.val:
+                # remove cur
+                nxt = cur.next
+                pre.next = nxt
+                
+                # insert cur to front
+                p1 = dummy
+                p2 = p1.next
+                while p2.val < cur.val:
+                    p1 = p1.next
+                    p2 = p2.next
+                cur.next = p2
+                p1.next = cur
+                
+                cur = nxt
+            else:
+                pre = cur
+                cur = cur.next
+        
+        return dummy.next