Update 371.sum-of-two-integers.md

problems/371.sum-of-two-integers.md

@@ -43,6 +43,8 @@ Output: 1
 - 求与之后左移一位来可以表示进位
 
 ## 代码
+代码支持：JS，C++，Java，Python  
+Javascript Code:
 ```js
 /*
  * @lc app=leetcode id=371 lang=javascript
@@ -62,4 +64,60 @@ var getSum = function(a, b) {
     return getSum(a ^ b, (a & b) << 1);
 };
 ```
+C++ Code:
+```c++
+class Solution {
+public:
+    int getSum(int a, int b) {
+        if(a==0) return b;
+        if(b==0) return a;
+
+        while(b!=0)
+        {
+            // 防止 AddressSanitizer 对有符号左移的溢出保护处理
+            auto carry = ((unsigned int ) (a & b))<<1;
+            // 计算无进位的结果
+            a = a^b;
+            //将存在进位的位置置1
+            b =carry;
+        }
+        return a;
+    }
+};
+```
 
+Java Code:
+```java
+class Solution {
+    public int getSum(int a, int b) {
+        if(a==0) return b;
+        if(b==0) return a;
+
+        while(b!=0)
+        {
+            int carry = a&b;
+            // 计算无进位的结果
+            a = a^b;
+            //将存在进位的位置置1
+            b =carry<<1;
+        }
+        return a;
+    }
+}
+```
+
+Python Code:
+```python
+# python整数类型为Unifying Long Integers, 即无限长整数类型.
+# 模拟 32bit 有符号整型加法
+class Solution:
+    def getSum(self, a: int, b: int) -> int:
+        a &= 0xFFFFFFFF
+        b &= 0xFFFFFFFF
+        while b:
+            carry = a & b
+            a ^= b
+            b = ((carry) << 1) & 0xFFFFFFFF
+            # print((a, b))
+        return a if a < 0x80000000 else ~(a^0xFFFFFFFF)
+```