[LeetCode] 190. Reverse Bits 颠倒二进制位

Reverse bits of a given 32 bits unsigned integer.

Example 1:

Input: 00000010100101000001111010011100
Output: 00111001011110000010100101000000
Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192 which its binary representation is 00111001011110000010100101000000.


Example 2:

Input: 11111111111111111111111111111101
Output: 10111111111111111111111111111111
Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471 which its binary representation is 10101111110010110010011101101001.

Note:

• Note that in some languages such as Java, there is no unsigned integer type. In this case, both input and output will be given as signed integer type and should not affect your implementation, as the internal binary representation of the integer is the same whether it is signed or unsigned.
• In Java, the compiler represents the signed integers using 2's complement notation. Therefore, in Example 2 above the input represents the signed integer -3 and the output represents the signed integer -1073741825.

If this function is called many times, how would you optimize it?

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
for (int i = 0; i < 32; ++i) {
if (n & 1 == 1) {
res = (res << 1) + 1;
} else {
res = res << 1;
}
n = n >> 1;
}
return res;
}
};

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
for (int i = 0; i < 32; ++i) {
res <<= 1;
if ((n & 1) == 1) ++res;
n >>= 1;
}
return res;
}
};

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
for (int i = 0; i < 32; ++i) {
res = (res << 1) | (n & 1);
n >>= 1;
}
return res;
}
};

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
for (int i = 0; i < 32; ++i) {
res = (res << 1) + (n >> i & 1);
}
return res;
}
};

class Solution {
public:
uint32_t reverseBits(uint32_t n) {
uint32_t res = 0;
for (int i = 0; i < 32; ++i) {
res |= ((n >> i) & 1) << (31 - i);
}
return res;
}
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/190

Number of 1 Bits

Reverse Integer

https://leetcode.com/problems/reverse-bits/

https://leetcode.com/problems/reverse-bits/discuss/54938/A-short-simple-Java-solution

https://leetcode.com/problems/reverse-bits/discuss/54772/The-concise-C++-solution(9ms)

https://leetcode.com/problems/reverse-bits/discuss/54741/O(1)-bit-operation-C++-solution-(8ms)

https://leetcode.com/problems/reverse-bits/discuss/54738/Sharing-my-2ms-Java-Solution-with-Explanation

https://leetcode.com/problems/reverse-bits/discuss/54873/Java-two-methods-using-String-or-bit-operation-6ms-and-2ms-easy-understand

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2015-03-08 10:15  Grandyang  阅读(23559)  评论(5编辑  收藏  举报