7. Reverse Integer

Problem

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 – 1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

 

Example 1:

Input: x = 123
Output: 321

Example 2:

Input: x = -123
Output: -321

Example 3:

Input: x = 120
Output: 21

Example 4:

Input: x = 0
Output: 0

Constraints:

• -231 <= x <= 231 – 1

Solution

function reverse(x: number): number {
    if(x==0) return 0; 
     if(x<0){
        x=-x;//去负号
        return -calculate(x);
    }
    if(x>0){
       return calculate(x);
    }
};
function calculate(x:number):number{
        var sum = 0;
        var s =x.toString();
        var l= s.length;
        for(let i =l-1;i>-1;i--){           
            let c = s.charAt(i);//长度是l,最后一个字符的序号是l-1
            let e = 10 ** i;//指数就是这个数在字符串里的序号
            let n= +c;//+n将字符串转化为数字
            sum += n * e;
        }
        if(sum>2**31-1 ||sum<-2**31) sum=0;
        return sum;
}

心得

目前水平比较渣渣所以先做简单的题……
这题是让把一个整数反转过来，思路就是把一个数倒过来其实就是把这个数的位数从后往前对应。然后再看这一位数字对应的需要乘以多少个10即可。

来源：

力扣（LeetCode）
链接：https://leetcode-cn.com/problems/reverse-integer
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