问题:计算两个数字的和?
分析:众所周知,Number类型的值有个上限,如果是两个极大的数字相加求和,便不能使用最简单的+方法,可以将两个数字当成字符串去处理,从最末尾开始相加,注意递进的位数,最后得到相加的字符串
实现:
//参数a、b均为字符串正整数,求和
function add(a, b){
var aLen = a.length,
bLen = b.length,
maxLen = aLen - bLen >= 0 ? aLen : bLen,
num = 0, //向上位数递级
str = '';
for(var i = 1; i <= maxLen; i++){
var aVal = aLen - i >= 0 ? parseInt(a.charAt(aLen - i)) : 0,
bVal = bLen - i >= 0 ? parseInt(b.charAt(bLen - i)) : 0,
bitVal = 0,
addVal = aVal + bVal + num;
if(addVal >= 10){
num = parseInt(addVal / 10);
bitVal = addVal % 10;
}else{
num = 0;
bitVal = addVal;
}
str += bitVal;
}
return Array.prototype.reduceRight.call(str, function(a, b){return a + b;});
}
第二种方式,递归算法
function add(a, b){
var aArr = a.split('').reverse(),
bArr = b.split('').reverse();
return (function addStep(aArr, bArr, str, num){
var aVal = aArr.length === 0 ? 0 : parseInt(aArr.shift()),
bVal = bArr.length === 0 ? 0 : parseInt(bArr.shift()),
num = num || 0,
bitVal = 0,
addVal = aVal + bVal + num,
str = str || '';
if(addVal >= 10){
num = parseInt(addVal / 10);
bitVal = addVal % 10;
}else{
num = 0;
bitVal = addVal;
}
str += bitVal;
if(aArr.length === 0 && bArr.length === 0){
console.log(Array.prototype.reduceRight.call(str, function(a, b){return a + b;}));
}else{
addStep(aArr, bArr, str, num);
}
})(aArr, bArr, '', 0)
}