算法之路(2) -- Reverse Integer leetcode第7题

今天这道是leetcode的第七题,难度仍是属于easy.如今我先挑着那些简单的作了,后面再作那些复杂滴.
题目是这样的:
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.

第一种方法是很直白的按照它的思路写下来:
var reverse = function(x) {

let sign = x>0 ? 1 : -1; 
let str = x + "";
let i = str.length-1;
while (str[i]=="0" && i!=0) {
    i--;
}

let reverseStr = str.substring(0, i+1).split('').reverse().join('');
   
let reverseX = parseInt(reverseStr);
if (reverseX > 2147483647 || reverseX < -2147483647)
        return 0;

return sign*reverseX;

};
最后是能够经过的,可是耗费的时间不少,上面那种方法有挺多须要改进的地方
如今换另外一种方法.

var reverse = function(x) {
    let reversed = 0;
    while(x) {
        reversed = reversed * 10 + x % 10;
        if (reversed > 2147483647 || reversed < -2147483647)
            return 0;
        x = parseInt(x/10);
    }
    return reversed;
};

这是最优的方法,上面那种方法定义变量用let比较好,最初我用的var,后来运行不如let快,别看小小一个地方,let比var快了10%呢,很可观的进步了.