LintCode 236: Swap Bits (位运算经典题)
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2022-07-05 12:59:04
...
我之前的做法太繁琐。
class Solution {
public:
/*
* @param x: An integer
* @return: An integer
*/
int swapOddEvenBits(int x) {
int len = sizeof(int) * 8;
for (int i = 0; i < len; i += 2) {
int even = (x >> i) & 0x1;
int odd = (x >> i + 1) & 0x1;
if (odd != even) {
if (even == 1) {
x &= ~(0x1 << i); //clear i th bit
x |= (0x1 << (i + 1)); //set the i+1 th bit
} else {
x &= ~(0x1 << (i + 1)); //clear i+1 th bit
x |= (0x1 << i); //set the i th bit
}
}
}
return x;
}
};
发现最简单的做法就是:
int swapOddEvenBits(int x) {
return ((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1);
}