[leetcode][javascript]Reverse Integer

最新推荐文章于 2025-05-05 14:12:43 发布

linc101

最新推荐文章于 2025-05-05 14:12:43 发布

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分类专栏： JavaScript on Leetcode 文章标签： javascript leetcode

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本文介绍了一种用于反转整数的算法实现，提供了两种方法：一是将整数转换为字符串并通过索引访问；二是通过除以10取余的方式。同时讨论了特殊情况处理，如输入整数最后一位为0的情况及如何避免反转后的整数溢出。

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Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

click to show spoilers.

Have you thought about this?

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.

主循环中写了两种办法来遍历数字：1.数字转化为字符串通过下标访问，2.通过除十取余

/**
 * @param {number} x
 * @return {number}
 */
var reverse = function(x) {
	var flag = x>0?1:-1;
    x = Math.abs(x);
	x +='';
	var res = 0;
	var t = 1;
	for(var i=0,l=x.length;i<l;i++){
		res = res + (x.charCodeAt(i)-48)*t;
		t*=10;
	}
	if(res>2147483647){
	    return 0;
	}
    return res*flag;
};

转换成字符串下标访问，通过unicode获取数值，不用再转回来，朦胧中，感觉这个快一点。（对leetcode的运行时间完全是醉了，完全不具参考性）

/**
 * @param {number} x
 * @return {number}
 */
var reverse = function(x) {
	var flag = x>0?1:-1;
    x = Math.abs(x);
	var res = 0;
	var t = 1;
	while(x){
		res = res*10 + x%10;
		x = Math.floor(x/10);
	}	
	if(res>2147483647){
	    return 0;
	}
    return res*flag;
};

是的个人更加喜欢前一种，and 偏爱for循环多一点。。。