add binary

add binary

Author: xczhang07

SUMMARY.md

@@ -43,7 +43,7 @@
    * [Gas Station](greedy/gas_station.md)
    * [Candy](greedy/candy.md)
    * [Word Break ](greedy/word_break.md)
-* [Linked List](linked_list/README)
+* [Linked List](linked_list/README.md)
    * [Linked List Cycle](linked_list/linked_list_cycle.md)
    * [Remove Duplicates from Sorted List ](linked_list/remove_duplicates_from_sorted_list.md)
    * [Merge Sorted Lists](linked_list/merge_sorted_lists.md)
@@ -57,4 +57,6 @@
    * [Copy List with Random Pointer](linked_list/copy_list_with_random_pointer.md)
 * [Math](math/README.md)
    * [Reverse Integer](math/reverse_integer.md)
+* [String](string/README.md)
+   * [Add Binary](string/add_binary.md)
 

linked_list/README.md

@@ -0,0 +1 @@
+# Linked List

string/README.md

@@ -0,0 +1,2 @@
+# String
+> 在这一章，我们将会覆盖leetcode上跟string有关联的题目.

string/add_binary.md

@@ -0,0 +1,82 @@
+# Add Binary
+
+> Given two binary strings, return their sum (also a binary string).
+
+> For example,
+a = "11"
+b = "1"
+Return "100".
+
+> 题目翻译: 对于给定的两个二进制数字所表达的字符串，我们求其相加所得到的结果， 根据上例便可得到答案.
+
+> 题目分析: 我认为这道题所要注意的地方涵盖以下几个方面:
+
+1. 对字符串的操作.
+2. 对于加法，我们应该建立一个进位单位，保存进位数值.
+3. 我们还要考虑两个字符串如果不同长度会怎样.
+4. int 类型和char类型的相互转换.
+
+> 时间复杂度:其实这就是针对两个字符串加起来跑一遍，O(n) n代表长的那串字符串长度.
+
+代码如下:
+
+```c++
+class Solution {
+public:
+    string addBinary(string a, string b) {
+        int len1 = a.size();
+        int len2 = b.size();
+        if(len1 == 0)
+            return b;
+        if(len2 == 0)
+            return a;
+
+        string ret;
+        int carry = 0;
+        int index1 = len1-1;
+        int index2 = len2-1;
+
+        while(index1 >=0 && index2 >= 0)
+        {
+            int num = (a[index1]-'0')+(b[index2]-'0')+carry;
+            carry = num/2;
+            num = num%2;
+            index1--;
+            index2--;
+            ret.insert(ret.begin(),num+'0');
+        }
+
+        if(index1 < 0&& index2 < 0)
+        {
+            if(carry == 1)
+            {
+                ret.insert(ret.begin(),carry+'0');
+                return ret;
+            }
+        }
+
+        while(index1 >= 0)
+        {
+            int num = (a[index1]-'0')+carry;
+            carry = num/2;
+            num = num%2;
+            index1--;
+            ret.insert(ret.begin(),num+'0');
+        }
+        while(index2 >= 0)
+        {
+            int num = (b[index2]-'0')+carry;
+            carry = num/2;
+            num = num%2;
+            index2--;
+            ret.insert(ret.begin(),num+'0');
+        }
+        if(carry == 1)
+            ret.insert(ret.begin(),carry+'0');
+        return ret;
+    }
+};
+
+```
+
+