Update 043._multiply_strings.md

Author: 0xkookoo

docs/Leetcode_Solutions/043._multiply_strings.md

@@ -7,35 +7,14 @@
 
 ## 难度: Medium 
 
-> 思路1
-
-虽然写了一堆similar problems，拿到的时候也算有思路，但是还是觉得很难写
 
 参考了别人的思路：
 
 1. m位的数字乘以n位的数字的结果最大为m+n位：
     * 999*99 < 1000*100 = 100000，最多为3+2 = 5位数。
 2. 先将字符串逆序便于从最低位开始计算。
 
-觉得这样写才是最容易理解的，看一个具体的🌰:
-
-```
-123 * 456
-
-	123
-  * 456
-
-先把每一位拿来相乘：得到
-			1   2   3
-		    4   5   6
-		    
-			6	12	18
-		5	10	15
-	4	8	12
 
-这样在全部加起来和做进位处理
-	5	6	0	8	8
-```
 
 ```python
 class Solution(object):
@@ -45,52 +24,31 @@ class Solution(object):
         :type num2: str
         :rtype: str
         """
-        if num1 == '0' or num2 == '0' : return '0'
-        len1,len2 = len(num1),len(num2)
-        
-        num1 = num1[::-1]
-        num2 = num2[::-1]
-        # 99 * 99 < 10000, maxmize 4 digit
-        arr = [0 for i in range(len1 + len2)]
-
-        for i in xrange(len1):
-            for j in xrange(len2):
-                arr[i+j] += (ord(num1[i]) - ord('0')) * (ord(num2[j]) - ord('0'))
-
-
-        res = [0 for i in range(len1 + len2)]
-
-        for i in range(len(arr)):
-            res[i] = arr[i] % 10
-            if i < len(arr) - 1:
-                arr[i+1] += arr[i]/10
+        lookup = {"0":0,"1":1,"2":2,"3":3,"4":4,"5":5,"6":6,"7":7,"8":8,"9":9} # 节省查找时间，避免无休止使用ord函数来得到数字
+        if num1 == '0' or num2 == '0':
+            return '0'
+        num1, num2 = num1[::-1], num2[::-1]
 
-        i = len(arr)-1
-        if res[i] == 0:
-            i -= 1
-        return ''.join(str(j) for j in res[:i+1][::-1])
+        tmp_res = [0 for i in range(len(num1)+len(num2))]
+        for i in range(len(num1)):
+            for j in range(len(num2)):
+                tmp_res[i+j] += lookup[num1[i]] * lookup[num2[j]]
+
+        res = [0 for i in range(len(num1)+len(num2))]
+        for i in range(len(num1)+len(num2)):
+            res[i] = tmp_res[i] % 10
+            if i < len(num1)+len(num2)-1:
+                tmp_res[i+1] += tmp_res[i]/10 
+        return ''.join(str(i) for i in res[::-1]).lstrip('0')   去掉最终结果头部可能存在的‘0’
 ```
 
-> 思路2:
-
-* 1.将字符串转化为数字
-* 2.然后相乘
-
-```python
-class Solution(object):
+觉得这样写才是最容易理解的，看一个具体的🌰:
+```
+input: num1, num2 = '91', '91'
+tmp_res = [1,18,81,0]
+res = [1,8,2,8]
 
-    def str2num(l):
-        if len(l)<=1:
-            return int(l[-1])
-        return 10**(len(l)-1) * int(l[0]) + Solution.str2num(l[1:])
+最终返回 "8281"
 
-    def multiply(self, num1, num2):
-        """
-        :type num1: str
-        :type num2: str
-        :rtype: str
-        """
-        n1 = Solution.str2num(num1)
-        n2 = Solution.str2num(num2)
-        return str(n1*n2)
+要注意最终返回头部可能会有‘0’，所以我们用lstrip去除一下
 ```