题目：两个非负大整数相加

解法1：在js中使用BigInt类型

解法2：

1const numLength = 15;
2
3function bigNumSum(bigNumA:string, bigNumB:string){
4    let resultA:Array<string> = [];
5    let resultB:Array<string> = [];
6    let result:Array<string> = []
7    let bigResult = bigNumA.length > bigNumB.length ? resultA :resultB;
8    for (let index = bigNumA.length; index > 0; index-=numLength) {
9        let start = index - numLength;
10        start = start >=0 ? start : 0;
11        resultA.push(bigNumA.substring(start, index));
12    }
13    for (let index = bigNumB.length; index > 0; index-=numLength) {
14        let start = index - numLength;
15        start = start >=0 ? start : 0;
16        resultB.push(bigNumB.substring(start, index));
17    }
18    let bigLocation:string = '0';
19    bigResult.forEach((_, index)=> {
20        let a:number = resultA[index]? Number(resultA[index]) : 0;
21        let b:number = resultB[index]? Number(resultB[index]) : 0;
22        let itemResult:string = String(a + b + Number(bigLocation));
23        if(itemResult.length > numLength) {
24            bigLocation = itemResult[0];
25            itemResult = itemResult.substring(1);
26        } else {
27            if(index !== bigResult.length -1 ){
28                itemResult = itemResult.padStart(numLength, '0')
29            }
30            bigLocation = '0'
31        }
32        result.unshift(itemResult)
33    })
34    bigLocation !== '0' &amp;&amp; result.unshift(bigLocation)
35    return result.join('')
36}
37

思路：

1. js中数字的最大精度范围为 Math.pow(2, 53) - 1 即 16 位，则设置分割长度 为 15位

2. 遍历原始数字从低位开始遍历，15位为一个元素 放入数组中

3. 挑选长度最大的数组遍历，下标相同的元素相加，设置进位变量，存储进位数值，相加后的结果放入结果数组