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Reverse bits of a given 32 bits unsigned integer.
For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as00111001011110000010100101000000).
Follow up:
If this function is called many times, how would you optimize it?
Related problem: Reverse Integer
两个指针i,j从前后向中间扫,如果bit[i] != bit[j],反转bit[i]和bit[j].
public class Solution {
// you need treat n as an unsigned value
public int reverseBits(int n) {
for (int i = 0; i < 16; i++) {
n = swapBit(n, i, 31-i);
}
return n;
}
public int swapBit(int n, int i, int j) {
int low = (n >> i) & 1;
int high = (n >> j) & 1;
if ((low ^ high) == 1) {
n ^= 1 << i;
n ^= 1 << j;
}
return n;
}
}
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原文地址:http://www.cnblogs.com/shini/p/4401345.html