Reverse Integer

Description

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output:  321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:

Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Tags: Math

思路0

题意是给你一个整型数，求它的逆序整型数，而且有个小坑点，当它的逆序整型数溢出的话，那么就返回 0，用我们代码表示的话可以求得结果保存在 long 中，最后把结果和整型的两个范围比较即可。

class Solution {
    public int reverse(int x) {
        long res = 0;
        for (; x != 0; x /= 10)
            res = res * 10 + x % 10;
        return res > Integer.MAX_VALUE || res < Integer.MIN_VALUE ? 0 : (int) res;
    }
}

思路1

记录正/负,通过字符串倒序,字符串转int,溢出必定会进入异常,则在catch处return 0即可。

class Solution {
    fun reverse(x: Int): Int {
        val symbol = if (x > 0) 1 else -1
        val absX = Math.abs(x)
        return try {
            Integer.parseInt(StringBuilder(String.format("%d", absX)).reverse().toString()) * symbol
        }catch(e: Exception) {
            0
        }
    }
}

var reverse = function(x) {
    var reverse = 0
    var arr = String(x).split('')
    if (x >= 0) {
	    reverse = Number(arr.reverse().join(''))
    } else {
	    reverse = -Number(arr.slice(1,arr.length).reverse().join(''))
    }
    if ((reverse >= Math.pow(2,31) - 1) || (reverse <= -Math.pow(2,31))) {
        reverse = 0
    }
    return reverse
};

结语

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