p190

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LeetCode p190 Reverse Bits 题解

1.题目：

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).

题意：

将一个数的二进制反转

2.解题思路:

位运算

3.代码

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public class Solution {
    // you need treat n as an unsigned value
    public int reverseBits(int n) {
        int ans = 0, count = 1;
		for (int i = 0; i < 31; i++) {
			ans += (1 & n);
			// System.out.println(ans);
			n = n >> 1;
			count = count * 2;
			ans = ans << 1;
		}
		ans += (1 & n);
		return ans;
    }
}

4.一些总结：