diff --git a/dynamicProgramming/last-stone-weight.py b/dynamicProgramming/last-stone-weight.py
new file mode 100644
index 0000000..a3798b9
--- /dev/null
+++ b/dynamicProgramming/last-stone-weight.py
@@ -0,0 +1,32 @@
+#coding: utf-8
+''' mbinary
+#######################################################################
+# File : last-stone-weight.py
+# Author: mbinary
+# Mail: zhuheqin1@gmail.com
+# Blog: https://mbinary.xyz
+# Github: https://github.com/mbinary
+# Created Time: 2019-05-28  23:30
+# Description:
+leetcode 1049: https://leetcode-cn.com/problems/last-stone-weight-ii/
+
+有一堆石头，每块石头的重量都是正整数。
+
+每一回合，从中选出任意两块石头，然后将它们一起粉碎。假设石头的重量分别为 x 和 y，且 x <= y。那么粉碎的可能结果如下：
+
+如果 x == y，那么两块石头都会被完全粉碎；
+如果 x != y，那么重量为 x 的石头将会完全粉碎，而重量为 y 的石头新重量为 y-x。
+最后，最多只会剩下一块石头。返回此石头最小的可能重量。如果没有石头剩下，就返回 0。
+
+#######################################################################
+'''
+
+class Solution:
+    def lastStoneWeightII(self, stones: List[int]) -> int:
+        sm = sum(stones)
+        ans = sm//2
+        dp = [0]*(ans+1)
+        for x in stones:
+            for j in range(ans,x-1,-1):
+                dp[j] = max(dp[j],dp[j-x]+x)
+    return sm-2*dp[ans]
diff --git a/dynamicProgramming/max-len-of-repeated-subarray.py b/dynamicProgramming/max-len-of-repeated-subarray.py
new file mode 100644
index 0000000..7ce0fcc
--- /dev/null
+++ b/dynamicProgramming/max-len-of-repeated-subarray.py
@@ -0,0 +1,21 @@
+#coding: utf-8
+''' mbinary
+#######################################################################
+# File : max-len-of-repeated-subarray.py
+# Author: mbinary
+# Mail: zhuheqin1@gmail.com
+# Blog: https://mbinary.xyz
+# Github: https://github.com/mbinary
+# Created Time: 2019-05-27  08:25
+# Description: 
+    给两个整数数组 A 和 B ，返回两个数组中公共的、长度最长的子数组的长度。
+#######################################################################
+'''
+def findLength(A,B):
+    n,m = len(A),len(B)
+    dp = [[0]*(m+1) for i in range(n+1)]
+    for i in range(1,n+1):
+        for j in range(1,m+1):
+            if A[i-1]==B[j-1]:
+                dp[i][j]=dp[i-1][j-1]+1
+    return max(max(row) for row in dp)
diff --git a/graph/dfs.py b/graph/dfs.py
new file mode 100644
index 0000000..6f45a84
--- /dev/null
+++ b/graph/dfs.py
@@ -0,0 +1,70 @@
+#coding: utf-8
+''' mbinary
+#######################################################################
+# File : dfs.py
+# Author: mbinary
+# Mail: zhuheqin1@gmail.com
+# Blog: https://mbinary.xyz
+# Github: https://github.com/mbinary
+# Created Time: 2019-05-27  10:02
+# Description: 
+    from leetcode-cn #1048: https://leetcode-cn.com/problems/longest-string-chain/
+    给出一个单词列表，其中每个单词都由小写英文字母组成。
+
+    如果我们可以在 word1 的任何地方添加一个字母使其变成 word2，那么我们认为 word1 是 word2 的前身。例如，"abc" 是 "abac" 的前身。
+
+    词链是单词 [word_1, word_2, ..., word_k] 组成的序列，k >= 1，其中 word_1 是 word_2 的前身，word_2 是 word_3 的前身，依此类推。
+
+    从给定单词列表 words 中选择单词组成词链，返回词链的最长可能长度。
+    
+#######################################################################
+'''
+
+class Solution:
+    def longestStrChain(self, words: List[str]) -> int:
+        def isAdj(s1,s2):
+            if len(s1)>len(s2):
+                s1,s2 = s2,s1
+            n1,n2 = len(s1),len(s2)
+            if n2-n1!=1:
+                return False
+            i=j=0
+            flag = False
+            while i<n1 and j<n2:
+                if s1[i]!=s2[j]:
+                    if flag:
+                        return False
+                    flag = True
+                    j+=1
+                else:
+                    i+=1
+                    j+=1
+            return True
+                        
+        def dfs(begin):
+            ans = 1
+            w = words[begin]
+            n = len(w)
+            if n+1 in lenDic:
+                for nd in lenDic[n+1]:
+                    #print(w,words[nd],isAdj(w,words[nd]))
+                    if isAdj(w,words[nd]):
+                        ans = max(ans,1+dfs(nd))
+            return ans
+        lenDic = {}
+        for i in range(len(words)):
+            n = len(words[i])
+            if n in lenDic:
+                lenDic[n].add(i)
+            else:
+                lenDic[n]={i}
+        
+        lens = sorted(lenDic)
+        n = len(lens)
+        ans = 0
+        for i in range(n):
+            if ans < n-i:
+                for nd in lenDic[lens[i]]:
+                    ans = max(ans,dfs(nd))          
+        return ans
+                
diff --git a/math/numWeight/addNegaBin.py b/math/numWeight/addNegaBin.py
new file mode 100644
index 0000000..d92e179
--- /dev/null
+++ b/math/numWeight/addNegaBin.py
@@ -0,0 +1,60 @@
+#coding: utf-8
+''' mbinary
+#######################################################################
+# File : addNegaBin.py
+# Author: mbinary
+# Mail: zhuheqin1@gmail.com
+# Blog: https://mbinary.xyz
+# Github: https://github.com/mbinary
+# Created Time: 2019-06-02  22:32
+# Description:
+
+给出基数为 -2 的两个数 arr1 和 arr2，返回两数相加的结果。
+
+数字以 数组形式 给出：数组由若干 0 和 1 组成，按最高有效位到最低有效位的顺序排列。例如，arr = [1,1,0,1] 表示数字 (-2)^3 + (-2)^2 + (-2)^0 = -3。数组形式 的数字也同样不含前导零：以 arr 为例，这意味着要么 arr == [0]，要么 arr[0] == 1。
+
+返回相同表示形式的 arr1 和 arr2 相加的结果。两数的表示形式为：不含前导零、由若干 0 和 1 组成的数组。
+
+eg
+输入：arr1 = [1,1,1,1,1], arr2 = [1,0,1]
+输出：[1,0,0,0,0]
+解释：arr1 表示 11，arr2 表示 5，输出表示 16
+#######################################################################
+'''
+from nega import nega
+
+def addNegaBin(arr1: list, arr2: list) -> list:
+    if len(arr1) < len(arr2):
+        arr1, arr2 = arr2, arr1
+    for i in range(-1, -len(arr2) - 1, -1):
+        if arr1[i] == 1 and arr2[i] == 1:
+            arr1[i] = 0
+            mux = 0
+            for j in range(i - 1, -len(arr1) - 1, -1):
+                if arr1[j] == mux:
+                    mux = 1 - mux
+                    arr1[j] = mux
+                else:
+                    arr1[j] = mux
+                    break
+            else:
+                arr1 = [1, 1] + arr1
+
+        elif arr1[i] == 0 and arr2[i] == 1:
+            arr1[i] = arr2[i]
+        #print(arr1,arr2,i)
+    while len(arr1) > 1 and arr1[0] == 0:
+        arr1.pop(0)
+    return arr1
+
+if __name__=='__main__':
+    while 1:
+        print("input q to quit or input x1 x2: ")
+        s = input()
+        if s=='q':
+            break
+        n1,n2 =[int(i) for i in  s.split()]
+        l1,l2 = nega(n1),nega(n2)
+        print(n1,l1)
+        print(n2,l2)
+        print(f'{n1}+{n2}={n1+n2}: {addNegaBin(l1,l2)}')
diff --git a/math/convertWeight.py b/math/numWeight/convertWeight.py
similarity index 100%
rename from math/convertWeight.py
rename to math/numWeight/convertWeight.py
diff --git a/math/numWeight/nega.py b/math/numWeight/nega.py
new file mode 100644
index 0000000..9750510
--- /dev/null
+++ b/math/numWeight/nega.py
@@ -0,0 +1,12 @@
+def nega(n:int,base=-2:int)->:list:
+    '''return list of num, the first is the highest digit'''
+    if base>-2:
+        raise Exception(f"[Error]: invalid base: {base}, base should be no more than -2")
+    ans = []
+    while n:
+        k = n%base
+        if k<0:
+            k-=base
+        ans.append(k)
+        n = (n-k)//base
+    return ans[::-1]