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This article mainly introduces java how to achieve integer inversion, has a certain reference value, interested friends can refer to, I hope you can learn a lot after reading this article, the following let Xiaobian take you to understand it.

Topic description

Given a 32-bit signed integer, you need to reverse the number on each of the integer.

Example 1:

Input: 123 output: 321

Example 2:

Input:-123 output:-321

Example 3:

Input: 120 output: 21

Note:

Assuming that our environment can only store 32-bit signed integers, the range of values is [− 231,231 − 1]. Based on this assumption, 0 is returned if the integer overflows after the inversion.

The idea of solving the problem

Tags: math

This question is very simple if the spillover problem is not taken into account. There are two ways to solve the overflow problem, the first is to solve the problem through string conversion plus try catch, and the second is to solve it through mathematical calculation.

Because the efficiency of string conversion is low and more library functions are used, the problem-solving scheme does not consider this method, but is solved by mathematical calculation.

Each bit of the number x is taken apart by a loop, and each step is determined whether or not to overflow when calculating the new value.

There are two overflow conditions, one is greater than the integer maximum value MAX_VALUE, the other is less than the integer minimum value MIN_VALUE, let the current calculation result is ans, the next bit is pop.

Judging from the overflow condition of ans * 10 + pop > MAX_VALUE

When ans > MAX_VALUE / 10 appears and there is still pop to be added, it must overflow

When ans = = MAX_VALUE / 10 and pop > 7, it must overflow. 7 is the single digit of 2 ^ 31-1.

From ans * 10 + pop

< MIN_VALUE这个溢出条件来看 当出现 ans < MIN_VALUE / 10 且 还有pop需要添加 时，则一定溢出 当出现 ans == MAX_VALUE / 10 且 pop < -8 时，则一定溢出，8是-2^31的个位数 代码class Solution { public int reverse(int x) { int ans = 0; while (x != 0) { int pop = x % 10; if (ans >

Integer.MAX_VALUE / 10 | (ans = = Integer.MAX_VALUE / 10 & & pop > 7) return 0; if (ans < Integer.MIN_VALUE / 10 | | (ans = = Integer.MIN_VALUE / 10 & & pop <-8) return 0; ans = ans * 10 + pop; x / = 10;} return ans;}}

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