publicintreverseBits3(intn){StringinputBinary=this.decimalToBinary(n);StringreversedInputBinary=newStringBuilder(inputBinary).reverse().toString();returnthis.binaryToDecimal(reversedInputBinary); }privateStringdecimalToBinary(intn){StringBuildersb=newStringBuilder();while(Integer.compareUnsigned(n,0) >0) ...
Java标准的Integer.reverse()源码。 代码 public class Solution { // you need treat n as an unsigned value public int reverseBits(int i) { i = (i & 0x55555555) << 1 | (i >>> 1) & 0x55555555; i = (i & 0x33333333) << 2 | (i >>> 2) & 0x33333333; i = (i & 0x0f0...
LeetCode 190. Reverse Bits (反转位) Reverse bits of a given 32 bits unsigned integer. For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000). 题目标签: Bit Manipulation 这道题目让我们把n的...
【Leetcode】Reverse Bits https://leetcode.com/problems/reverse-bits/ 题目: Reverse bits of a given 32 bits unsigned integer. 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000). Follow up: If this function is called many times, how would you op...
LeetCode Top Interview Questions 190. Reverse Bits (Java版; Easy) 题目描述 Reverse bits of a given 32 bits unsigned integer. Example 1: Input: 00000010100101000001111010011100 Output: 00111001011110000010100101000000 Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned inte...
LeetCode#190: Reverse Bits Description Reverse bits of a given 32 bits unsigned integer. Example Solution 我们把要翻转的数n从低位到高位依次放入res中,放入的方式就是判断此时n的最低位是0还是1,如果是1就将res++,否则直接无视就好了。放入后我们把n右移一位,使高一位成为最低位以做下一次的判断,而...
Reverse bits of a given 32 bits unsigned integer. Example 1: Input: 00000010100101000001111010011100 Output: 00111001011110000010100101000000 Explanation: The input binary string 000000101001010000011... 查看原文 【190-Reverse Bits(反转二制)】 )】【LeetCode-面试算法经典-Java实现】【所有题目目录索引】 代码...
classSolution {public: uint32_t reverseBits(uint32_t n) { uint32_t res=0;for(inti =0; i <32; ++i) { res|= ((n >> i) &1) << (31-i); }returnres; } }; 本文转自博客园Grandyang的博客,原文链接:翻转位[LeetCode] Reverse Bits,如需转载请自行联系原博主。
Time Complexity: O(N), where N is number of bits in the input integer Space Complexity: O(1), no extra space needed References Leetcode discussion solution
颠倒给定的 32 位无符号整数的二进制位。 提示: 请注意,在某些语言(如 Java)中,没有无符号整数类型。在这种情况下,输入和输出都将被指定为有符号整数类型,并且不应影响您的实现,因为无论整数是有符号的还是无符号的,其内部的二进制表示形式都是相同的。