剑指 Offer 65. 不用加减乘除做加法
题目描述：
写一个函数，求两个整数之和，要求在函数体内不得使用 “+”、“-”、“*”、“/” 四则运算符号。
示例:
输入: a = 1, b = 1
输出: 2
提示：
a, b 均可能是负数或 0
结果不会溢出 32 位整数
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/bu-yong-jia-jian-cheng-chu-zuo-jia-fa-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
分析：我真的觉得这道题不应该作为简单题出现。。。
熟悉位运算一眼就会，不熟悉的死活做不出来。我个人的理解只能想到无进位和通过异或得出，进位通过与运算得出。但是如何处理连环进位的问题我就不知道了。
只好去看了题解，真的再一次叹服于大佬的智慧。
来源：leetcode用户 Krahets
https://leetcode-cn.com/problems/bu-yong-jia-jian-cheng-chu-zuo-jia-fa-lcof/solution/mian-shi-ti-65-bu-yong-jia-jian-cheng-chu-zuo-ji-7/
主要是那个循环求n和c， 一般情况下真的想不到。。。。
以及负数的问题，这篇文章并没有提到，一输入负数就报错，看了看评论区，把其中一句改成：
int c=(unsigned int)(a&b)<<1;
即可，具体原因稍后分析。
代码附上：
class Solution {
public:
int add(int a, int b) {
while(b!=0)
{
int c=(unsigned int)(a&b)<<1;
a=a^b;
b=c;
}
return a|b;
}
};要探讨加上unsigned int这个问题，又必须要去讨论数值在计算机中的原码，反码和补码的问题了。
这篇文章讨论的很好https://www.dotcpp.com/wp/221.html，这里我就不再赘述了。
然后是为什么要加上unsigned int呢
我们先看看，如果不加unsigned int，每次循环的进位和无进位和是多少
jinwei is 2 he is -2
jinwei is 4 he is -4
jinwei is 8 he is -8
jinwei is 16 he is -16
jinwei is 32 he is -32
jinwei is 64 he is -64
jinwei is 128 he is -128
jinwei is 256 he is -256
jinwei is 512 he is -512
jinwei is 1024 he is -1024
jinwei is 2048 he is -2048
jinwei is 4096 he is -4096
jinwei is 8192 he is -8192
jinwei is 16384 he is -16384
jinwei is 32768 he is -32768
jinwei is 65536 he is -65536
jinwei is 131072 he is -131072
jinwei is 262144 he is -262144
jinwei is 524288 he is -524288
jinwei is 1048576 he is -1048576
jinwei is 2097152 he is -2097152
jinwei is 4194304 he is -4194304
jinwei is 8388608 he is -8388608
jinwei is 16777216 he is -16777216
jinwei is 33554432 he is -33554432
jinwei is 67108864 he is -67108864
jinwei is 134217728 he is -134217728
jinwei is 268435456 he is -268435456
jinwei is 536870912 he is -536870912
jinwei is 1073741824 he is -1073741824
jinwei is -2147483648 he is -2147483648
当进位为 -2147483648的时候，下一步进行左移操作，就会导致进位溢出，报错，然后无法继续。
而无符号整数是不存在溢出这个问题的，详见这篇文章：https://blog.csdn.net/gongchangwangqi/article/details/47210447
故当使用unsigned int时， -2147483648左移一位，变为了0，并且不会报溢出错误。因此，循环结束，可以计算出结果。
当加上unsigned int之后的结果：
jinwei is 2 he is -2
jinwei is 4 he is -4
jinwei is 8 he is -8
jinwei is 16 he is -16
jinwei is 32 he is -32
jinwei is 64 he is -64
jinwei is 128 he is -128
jinwei is 256 he is -256
jinwei is 512 he is -512
jinwei is 1024 he is -1024
jinwei is 2048 he is -2048
jinwei is 4096 he is -4096
jinwei is 8192 he is -8192
jinwei is 16384 he is -16384
jinwei is 32768 he is -32768
jinwei is 65536 he is -65536
jinwei is 131072 he is -131072
jinwei is 262144 he is -262144
jinwei is 524288 he is -524288
jinwei is 1048576 he is -1048576
jinwei is 2097152 he is -2097152
jinwei is 4194304 he is -4194304
jinwei is 8388608 he is -8388608
jinwei is 16777216 he is -16777216
jinwei is 33554432 he is -33554432
jinwei is 67108864 he is -67108864
jinwei is 134217728 he is -134217728
jinwei is 268435456 he is -268435456
jinwei is 536870912 he is -536870912
jinwei is 1073741824 he is -1073741824
jinwei is -2147483648 he is -2147483648
jinwei is 0 he is 0