【Codewars】<4kyu>Sum Strings as Numbers
阿新 • • 發佈:2020-12-31
技術標籤:Codewars演算法字串jsjavascript前端
題目
<4kyu>Sum Strings as Numbers
Given the string representations of two integers, return the string representation of the sum of those integers.
A string representation of an integer will contain no characters besides the ten numerals “0” to “9”.
給定兩個整數的字串表示形式,返回這些整數之和的字串表示形式。
(注:這兩個字串將不包含除“0”到“9”之外的任何字元。)
例子
sumStrings('1','2') // => '3'
sumStrings('','5') // => '5'
sumStrings('001','5') // => '6'
sumStrings('50095301248058391139327916261','81055900096023504197206408605') // => '131151201344081895336534324866'
題解一
// 題解一:
function sumStrings(a,b) {
var result = '';
var remainder = 0;
if(a.length < b.length) {
var c = a;
a = b;
b = c;
}
for(let i=1;i<=a.length;i++){
// 從最右邊開始,將兩個數求和,並加上上一次的餘數,首次餘數為0
var sum = (a.length-(i-1)>0 ? parseInt(a.substr(a.length-i,1)) : 0) + (b.length-(i-1)>0 ? parseInt(b.substr(b.length-i,1) ) : 0) + remainder;
// 求和後去取餘,就是該位要保留的數字
result = sum%10 + result
// 計算餘數
remainder = parseInt(sum/10)
// 如果到了最左邊,即a的第一位,餘數還大於0,則直接加上餘數
if(remainder > 0 && i == a.length) result = remainder + result;
}
return result.replace(/\b(0+)/gi,""); // 去除字串前面的0
}
題解二(Best Practices)
// 題解二:
function sumStrings(a, b) {
var res = '', c = 0;
a = a.split('');
b = b.split('');
while (a.length || b.length || c) {
c += ~~a.pop() + ~~b.pop();
res = c % 10 + res;
c = c > 9;
}
return res.replace(/^0+/, '');
}
小夥伴們有其它更好的解法,歡迎評論區提出交流~