diff --git a/Algorithms/125. Valid-Palindrome/README.md b/Algorithms/125. Valid-Palindrome/README.md
index 9ddcff27..7c97f7db 100644
--- a/Algorithms/125. Valid-Palindrome/README.md	
+++ b/Algorithms/125. Valid-Palindrome/README.md	
@@ -5,10 +5,14 @@
 Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
 
 For example,
+
+```c
 "A man, a plan, a canal: Panama" is a palindrome.
 "race a car" is not a palindrome.
+```
+
+Note:  
 
-Note:
 Have you consider that the string might be empty? This is a good question to ask during an interview.
 
 For the purpose of this problem, we define empty string as valid palindrome.
diff --git a/Algorithms/141. Linked List Cycle/README.md b/Algorithms/141. Linked List Cycle/README.md
index 25adbe52..e76c87da 100644
--- a/Algorithms/141. Linked List Cycle/README.md	
+++ b/Algorithms/141. Linked List Cycle/README.md	
@@ -4,7 +4,7 @@
 
 Given a linked list, determine if it has a cycle in it.
 
-Follow up:
+Follow up:   
 Can you solve it without using extra space?
 
 
diff --git a/Algorithms/147. Insertion Sort List/README.md b/Algorithms/147. Insertion Sort List/README.md
index 74fdb3d3..8a5f4ebb 100644
--- a/Algorithms/147. Insertion Sort List/README.md	
+++ b/Algorithms/147. Insertion Sort List/README.md	
@@ -4,6 +4,7 @@
 
 Sort a linked list using insertion sort.
 
+![](https://upload.wikimedia.org/wikipedia/commons/0/0f/Insertion-sort-example-300px.gif)
 
 A graphical example of insertion sort. The partial sorted list (black) initially contains only the first element in the list.
 With each iteration one element (red) is removed from the input data and inserted in-place into the sorted list
diff --git a/Algorithms/160. Intersection of Two Linked Lists/README.md b/Algorithms/160. Intersection of Two Linked Lists/README.md
index bd53df67..9025e5db 100644
--- a/Algorithms/160. Intersection of Two Linked Lists/README.md	
+++ b/Algorithms/160. Intersection of Two Linked Lists/README.md	
@@ -6,11 +6,14 @@ Write a program to find the node at which the intersection of two singly linked
 
 For example, the following two linked lists:
 
+![](https://assets.leetcode.com/uploads/2018/12/13/160_statement.png)
 
 begin to intersect at node c1.
 
 Example 1:
 
+![](https://assets.leetcode.com/uploads/2018/12/13/160_example_1.png)
+
 ```c
 Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
 Output: Reference of the node with value = 8
@@ -19,6 +22,8 @@ Input Explanation: The intersected node's value is 8 (note that this must not be
 
 Example 2:
 
+![](https://assets.leetcode.com/uploads/2018/12/13/160_example_2.png)
+
 ```c
 Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
 Output: Reference of the node with value = 2
@@ -28,6 +33,8 @@ Input Explanation: The intersected node's value is 2 (note that this must not be
 
 Example 3:
 
+![](https://assets.leetcode.com/uploads/2018/12/13/160_example_3.png)
+
 ```c
 Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
 Output: null
diff --git a/Algorithms/41.First-Missing-Positive/First Missing Positive.go b/Algorithms/41. First-Missing-Positive/41. First Missing Positive.go
similarity index 100%
rename from Algorithms/41.First-Missing-Positive/First Missing Positive.go
rename to Algorithms/41. First-Missing-Positive/41. First Missing Positive.go
diff --git a/Algorithms/41.First-Missing-Positive/First Missing Positive_test.go b/Algorithms/41. First-Missing-Positive/41. First Missing Positive_test.go
similarity index 100%
rename from Algorithms/41.First-Missing-Positive/First Missing Positive_test.go
rename to Algorithms/41. First-Missing-Positive/41. First Missing Positive_test.go
diff --git a/Algorithms/41.First-Missing-Positive/README.md b/Algorithms/41. First-Missing-Positive/README.md
similarity index 100%
rename from Algorithms/41.First-Missing-Positive/README.md
rename to Algorithms/41. First-Missing-Positive/README.md
diff --git a/Algorithms/725. Split Linked List in Parts/725. Split Linked List in Parts.go b/Algorithms/725. Split Linked List in Parts/725. Split Linked List in Parts.go
new file mode 100644
index 00000000..34b803aa
--- /dev/null
+++ b/Algorithms/725. Split Linked List in Parts/725. Split Linked List in Parts.go	
@@ -0,0 +1,64 @@
+package leetcode
+
+import "fmt"
+
+/**
+ * Definition for singly-linked list.
+ * type ListNode struct {
+ *     Val int
+ *     Next *ListNode
+ * }
+ */
+func splitListToParts(root *ListNode, k int) []*ListNode {
+	res := make([]*ListNode, 0)
+	if root == nil {
+		for i := 0; i < k; i++ {
+			res = append(res, nil)
+		}
+		return res
+	}
+	length := getLength(root)
+	splitNum := length / k
+	lengNum := length % k
+	cur := root
+	head := root
+	pre := root
+	fmt.Printf("总长度 %v, 分 %v 组, 前面 %v 组长度为 %v, 剩余 %v 组，每组 %v\n", length, k, lengNum, splitNum+1, k-lengNum, splitNum)
+	if splitNum == 0 {
+		for i := 0; i < k; i++ {
+			if cur != nil {
+				pre = cur.Next
+				cur.Next = nil
+				res = append(res, cur)
+				cur = pre
+			} else {
+				res = append(res, nil)
+			}
+		}
+		return res
+	}
+	for i := 0; i < lengNum; i++ {
+		for j := 0; j < splitNum; j++ {
+			cur = cur.Next
+		}
+		fmt.Printf("0 刚刚出来 head = %v cur = %v pre = %v\n", head, cur, head)
+		pre = cur.Next
+		cur.Next = nil
+		res = append(res, head)
+		head = pre
+		cur = pre
+		fmt.Printf("0 head = %v cur = %v pre = %v\n", head, cur, head)
+	}
+	for i := 0; i < k-lengNum; i++ {
+		for j := 0; j < splitNum-1; j++ {
+			cur = cur.Next
+		}
+		fmt.Printf("1 刚刚出来 head = %v cur = %v pre = %v\n", head, cur, head)
+		pre = cur.Next
+		cur.Next = nil
+		res = append(res, head)
+		head = pre
+		cur = pre
+	}
+	return res
+}
diff --git a/Algorithms/725. Split Linked List in Parts/725. Split Linked List in Parts_test.go b/Algorithms/725. Split Linked List in Parts/725. Split Linked List in Parts_test.go
new file mode 100644
index 00000000..a27c893f
--- /dev/null
+++ b/Algorithms/725. Split Linked List in Parts/725. Split Linked List in Parts_test.go	
@@ -0,0 +1,87 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question725 struct {
+	para725
+	ans725
+}
+
+// para 是参数
+// one 代表第一个参数
+type para725 struct {
+	one []int
+	n   int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans725 struct {
+	one []int
+}
+
+func Test_Problem725(t *testing.T) {
+
+	qs := []question725{
+
+		question725{
+			para725{[]int{1, 2, 3, 4, 5}, 7},
+			ans725{[]int{1, 2, 3, 4, 5, 0, 0}},
+		},
+
+		question725{
+			para725{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 3},
+			ans725{[]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}},
+		},
+
+		question725{
+			para725{[]int{1, 1, 1, 1, 1}, 1},
+			ans725{[]int{1, 1, 1, 1, 1}},
+		},
+
+		question725{
+			para725{[]int{}, 3},
+			ans725{[]int{}},
+		},
+
+		// question725{
+		// 	para725{[]int{1, 2, 3, 2, 3, 2, 3, 2}, 0},
+		// 	ans725{[]int{1, 2, 3, 2, 3, 2, 3, 2}},
+		// },
+
+		// question725{
+		// 	para725{[]int{1, 2, 3, 4, 5}, 5},
+		// 	ans725{[]int{1, 2, 3, 4}},
+		// },
+
+		// question725{
+		// 	para725{[]int{}, 5},
+		// 	ans725{[]int{}},
+		// },
+
+		// question725{
+		// 	para725{[]int{1, 2, 3, 4, 5}, 10},
+		// 	ans725{[]int{1, 2, 3, 4, 5}},
+		// },
+
+		// question725{
+		// 	para725{[]int{1}, 1},
+		// 	ans725{[]int{}},
+		// },
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 725------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans725, q.para725
+		res := splitListToParts(S2l(p.one), p.n)
+		for _, value := range res {
+			fmt.Printf("【input】:%v    length:%v   【output】:%v\n", p, len(res), L2s(value))
+		}
+		fmt.Printf("\n\n\n")
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/725. Split Linked List in Parts/README.md b/Algorithms/725. Split Linked List in Parts/README.md
new file mode 100644
index 00000000..0952b379
--- /dev/null
+++ b/Algorithms/725. Split Linked List in Parts/README.md	
@@ -0,0 +1,55 @@
+# [725. Split Linked List in Parts](https://leetcode.com/problems/add-two-numbers-ii/)
+
+## 题目
+
+Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts".
+
+The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.
+
+The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.
+
+Return a List of ListNode's representing the linked list parts that are formed.
+
+Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]
+
+Example 1:
+
+```c
+Input: 
+root = [1, 2, 3], k = 5
+Output: [[1],[2],[3],[],[]]
+Explanation:
+The input and each element of the output are ListNodes, not arrays.
+For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.
+The first element output[0] has output[0].val = 1, output[0].next = null.
+The last element output[4] is null, but it's string representation as a ListNode is [].
+```
+
+Example 2:
+
+```c
+Input: 
+root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
+Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
+Explanation:
+The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.
+```
+
+Note:
+
+- The length of root will be in the range [0, 1000].
+- Each value of a node in the input will be an integer in the range [0, 999].
+- k will be an integer in the range [1, 50].
+
+
+
+## 题目大意
+
+把链表分成 K 个部分，要求这 K 个部分尽量两两长度相差不超过 1，并且长度尽量相同。
+
+把链表长度对 K 进行除法，结果就是最终每组的长度 n。把链表长度对 K 进行取余操作，得到的结果 m，代表前 m 组链表长度为 n + 1 。相当于把多出来的部分都分摊到前面 m 组链表中了。最终链表是前 m 组长度为 n + 1，后 K - m 组链表长度是 n。
+
+注意长度不足 K 的时候要用 nil 进行填充。
+
+
+