【[LeetCode] 190. Reverse Bits_Easy tag: Bit Manipulation】的更多相关文章

[LeetCode] 190. Reverse Bits_Easy tag: Bit Manipulation

Reverse bits of a given 32 bits unsigned integer. Example: Input: 43261596 Output: 964176192 Explanation: 43261596 represented in binary as 00000010100101000001111010011100,   return 964176192 represented in binary as 00111001011110000010100101000000…

LeetCode 190. Reverse Bits (算32次即可)

题目: 190. Reverse Bits 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). Follo…

[LeetCode] 190. Reverse Bits 翻转二进制位

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). Follow up:If this function…

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…

[LeetCode] 190. Reverse Bits 颠倒二进制位

Reverse bits of a given 32 bits unsigned integer. Example 1: Input: 00000010100101000001111010011100 Output: 00111001011110000010100101000000 Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 432615…

Java for LeetCode 190 Reverse Bits

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). JAVA实现如下: public int revers…

Leetcode 190. Reverse Bits（反转比特数）

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). Follow up: If this function…

Leetcode 190 Reverse Bits 位运算

反转二进制 class Solution { public: uint32_t reverseBits(uint32_t n) { uint32_t ans = ; ; i<; ++i,n >>=){ ans = (ans<< )|(n & ); } return ans; } };…

Java [Leetcode 190]Reverse Bits

题目描述: everse bits of a given 32 bits unsigned integer. For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000). 解题思路: 移位操作. 代码如下: publi…

[LeetCode] 476. Number Complement_Easy tag: Bit Manipulation

这个题目思路就是比如101 的结果是010, 可以从111^101 来得到, 那么我们就需要知道刚好比101多一位的1000, 所以利用 while i <= num : i <<= 1, 去得到1000, 然后-1, 便得到111, 再跟num ^, 也就是异或即可. Code class Solution(object): def findComplement(self, num): i = 1 while i <= num: i <<= 1 return (i-1…