[LeetCode 190] Reverse Bits

문제 바로가기

Description

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Reverse bits of a given 32 bits signed integer.

Example 1

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• Input: n = 43261596
• Output: 964176192
• Explanation:

Integer	Binary
43261596	00000010100101000001111010011100
964176192	00111001011110000010100101000000

Example 2

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• Input: n = 2147483644
• Output: 1073741822
• Explanation:

Integer	Binary
2147483644	01111111111111111111111111111100
1073741822	00111111111111111111111111111110

Constraints

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• 0 <= n <= 2^31 - 2
• n is even.

Code

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내 제출

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class Solution {
    public int reverseBits(int n) {
        int ans = 0;
        for (int i = 0; i < 32 && n != 0; ++i) {
            ans |= (n & 1) << (31 - i);
            n >>>= 1;
        }
        return ans;
    }
}

Runtime	Memory
0 ms	40.9 MB

다른 풀이

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class Solution {
    public int reverseBits(int n) {

        int ans=0;

        for(int i=0;i<32;i++){

            //first ateking last bit by and opearting then shidting the ans bit and doing or
            ans=(n&1) | ans<<1;
            //going to thennext bit
            n=n>>1;
        }
        return ans;
        
    }
}

Reference

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• https://github.com/doocs/leetcode/blob/main/solution/0100-0199/0190.Reverse%20Bits/Solution.java