java.lang.Integer.reverse()是Java中的一个内置方法,用于返回指定int值的二进制补码中比特的反向顺序。语法。public static int reverse(int a) Java Copy参数。参数a是一个整数值,其位数要被颠倒。返回值。该方法返回通过颠倒指定的int值中的位的顺序而得到的值。
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 store integers within the 32-bit signed integer range:...
int型的数值范围是 -2147483648～2147483647, 那么如果我们要翻转 1000000009 这个在范围内的数得到 9000000001,而翻转后的数就超过了范围 Java解法: classSolution {publicintreverse(intx) {longresult = 0;while(x != 0){ result= result * 10 + x % 10; x= x / 10; }if(result > Integer.MAX_VALUE...
reverse(int i) Returns the value obtained by reversing the order of the bits in the two's complement binary representation of the specified int value. static int reverseBytes(int i) Returns the value obtained by reversing the order of the bytes in the two's complement representation of the...
Java.Lang 組件: Mono.Android.dll 傳回值,這個值會反轉兩個補數二進位表示中指定int值的位順序。 C# [Android.Runtime.Register("reverse","(I)I","")]publicstaticintReverse(inti); 參數 i Int32 要反轉的值 傳回 Int32 藉由反轉指定int值中位的順序取得的值。
leetCode:reverseInteger 反向整数 【JAVA实现】 反向整数 给定一个 32 位有符号整数,将整数中的数字进行反转,如果超出整数的最大或者最小范围返回0 更多文章查看个人博客个人博客地址:反向整数 方法一 利用StringBuilder的reverse方法,将数字转换成字符反转然后再转换回整数...
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. 详见:https://leetcode.com/problems/reverse-integer/description/ ...
reverse嘛，就是把组成Integer的32个比特的顺序调个方向。假设是32个学生排队，原来是按学号1~32顺序...
To reverse an integer using awhileloop, we must follow all three steps mentioned. Example: importjava.util.Scanner;publicclassReverse_While{publicstaticvoidmain(String args[]){System.out.print("Enter the Integer you want to Reverse: ");Scanner input_num=newScanner(System.in);intinput_number=...
int reverse(int x) { if (x == INT_MIN) return ; bool isNeg = x < ; x = abs(x); long res = ; while (x) { res = res * + x % ; if (res > INT_MAX) return ; x /= ; } return isNeg ? -(int)res: (int)res; } }; 【JAVA、C++】LeetCode 007 Reverse Integer的更...