iOS中计算两个大数相加算法（OC实现）

Posted 2020-07-26

我们知道计算机的数据类型不同，所能表示的数据量级也不相同，比如：

unsigned int : 0～4294967295   
int : -2147483648～2147483647 
unsigned long : 0～4294967295
long :  -2147483648～2147483647
long long : -9223372036854775808 ~ 9223372036854775807
unsigned long long : 0 ~ 18446744073709551615

__int64 :  -9223372036854775808 ~ 9223372036854775807

unsigned __int64 : 0 ~ 18446744073709551615

日常的ios开发工作中使用的数据量级并不会超过上述数据类型，如果是两个大数（比如说：99999999999999999999999999999999999999甚至更多位数的），如何进行计算呢？

下面是我自己用OC写的两个大数相加的算法：

（基本思路是利用字符串模拟两个整数相加的过程）

//两个大数相加算法
-(NSString *)addTwoNumberWithOneNumStr:(NSString *)one anotherNumStr:(NSString *)another
{
    int i = 0;
    int j = 0;
    int maxLength = 0;
    int sum = 0;
    int overflow = 0;
    int carryBit = 0;
    NSString *temp1 = @"";
    NSString *temp2 = @"";
    NSString *sums = @"";
    NSString *tempSum = @"";
    int length1 = (int)one.length;
    int length2 = (int)another.length;
    //1.反转字符串
    for (i = length1 - 1; i >= 0 ; i--) {
        NSRange range = NSMakeRange(i, 1);
        temp1 = [temp1 stringByAppendingString:[one substringWithRange:range]];
        NSLog(@"%@",temp1);
    }
    for (j = length2 - 1; j >= 0; j--) {
        NSRange range = NSMakeRange(j, 1);
        temp2 = [temp2 stringByAppendingString:[another substringWithRange:range]];
        NSLog(@"%@",temp2);
    }
    
    //2.补全缺少位数为0
    maxLength = length1 > length2 ? length1 : length2;
    if (maxLength == length1) {
        for (i = length2; i < length1; i++) {
            temp2 = [temp2 stringByAppendingString:@"0"];
            NSLog(@"i = %d --%@",i,temp2);
        }
    }else{
        for (j = length1; j < length2; j++) {
            temp1 = [temp1 stringByAppendingString:@"0"];
            NSLog(@"j = %d --%@",j,temp1);
        }
    }
    //3.取数做加法
    for (i = 0; i < maxLength; i++) {
        NSRange range = NSMakeRange(i, 1);
        int a = [temp1 substringWithRange:range].intValue;
        int b = [temp2 substringWithRange:range].intValue;
        sum = a + b + carryBit;
        if (sum > 9) {
            if (i == maxLength -1) {
                overflow = 1;
            }
            carryBit = 1;
            sum -= 10;
        }else{
            carryBit = 0;
        }
        tempSum = [tempSum stringByAppendingString:[NSString stringWithFormat:@"%d",sum]];
    }
    if (overflow == 1) {
        tempSum = [tempSum stringByAppendingString:@"1"];
    }
    int sumlength = (int)tempSum.length;
    for (i = sumlength - 1; i >= 0 ; i--) {
        NSRange range = NSMakeRange(i, 1);
        sums = [sums stringByAppendingString:[tempSum substringWithRange:range]];
    }
    NSLog(@"sums = %@",sums);
    return sums;
}

算法写的可能不是很简洁，如果你有更好的写法，欢迎在评论区提出，以上所有内容均为本人手打，如需转载请注明出处！