给定一个无符号整数, 求2进制翻转后的结果.
Input: 00000010100101000001111010011100
Output: 00111001011110000010100101000000
// intput: 43261596 output: 964176192
将整数分解为2的i次方的和, 如果含有2的i次方, 结果就增加2的(32-i-1)次方.
uint32_t reverseBits(uint32_t n) {
int res = 0;
int i = 0;
while(n > 0) {
int x = 1<<i++;
if ((n & x) == x) {
n -= x;
res += 1<<(32-i);
}
}
return res;
}
位运算的运用.
题目 给定一个无符号整数, 求2进制翻转后的结果. 思路 将整数分解为2的i次方的和, 如果含有2的i次方, 结果...
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本文标题：Leetcode.190.Reverse Bits
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