问题
I'm currently stuck on a Codewars challenge that I can't get my head around:
Given a string representation of two integers, return the string representation of those integers, e.g. sumStrings('1','2') // => '3'
I've used the following code so far, but it fails on large number test cases as the number is converted into a scientific notation:
function sumStrings(a,b) {
var res = +a + +b;
return res.toString();
}
Any help would be much appreciated.
Edit:
Fiddle example: https://jsfiddle.net/ag1z4x7d/
回答1:
The problem is that in that specific kata (IIRC), the numbers stored in a
and b
are too large for a regular 32 bit integer, and floating point arithmetic isn't exact. Therefore, your version does not return the correct value:
sumStrings('100000000000000000000', '1')
// returns '100000000000000000000' instead of '100000000000000000001'
You have to make sure that this does not happen. One way is to do an good old-fashioned carry-based addition and stay in the digit/character based world throughout the whole computation:
function sumStrings(a, b) {
var digits_a = a.split('')
var digits_b = b.split('')
...
}
回答2:
function sumStrings(a, b) { // sum for any length
function carry(value, index) { // cash & carry
if (!value) { // no value no fun
return; // leave shop
}
this[index] = (this[index] || 0) + value; // add value
if (this[index] > 9) { // carry necessary?
carry.bind(this)(this[index] / 10 | 0, index + 1); // better know this & go on
this[index] %= 10; // remind me later
}
}
var array1 = a.split('').map(Number).reverse(), // split stuff and reverse
array2 = b.split('').map(Number).reverse(); // here as well
array1.forEach(carry, array2); // loop baby, shop every item
return array2.reverse().join(''); // return right ordered sum
}
document.write(sumStrings('999', '9') + '<br>');
document.write(sumStrings('9', '999') + '<br>');
document.write(sumStrings('1', '9999999999999999999999999999999999999999999999999999') + '<br>');
来源:https://stackoverflow.com/questions/36080824/sum-big-integers