leetcode:Reverse Bits
2015-07-19 11:05
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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
分析:题意为反转给定32位无符号整型数的位
思路:我们只需将原整型数从右到左一个一个取出来,然后一个个加到新数的最低位中即可
或可参考更简洁做法:
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;
}
};
其他可参考方法:
或:
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
分析:题意为反转给定32位无符号整型数的位
思路:我们只需将原整型数从右到左一个一个取出来,然后一个个加到新数的最低位中即可
class Solution { public: uint32_t reverseBits(uint32_t n) { uint32_t s=0; for(int i=0;i<32;i++){ if(n&1==1){ n>>=1; s=(s<<=1)+1; } else{ n>>=1; s=(s<<=1); } } return s; } };
或可参考更简洁做法:
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;
}
};
其他可参考方法:
#include<iostream> using namespace std; class Solution { public: uint32_t reverseBits(uint32_t n) { uint32_t m=0; for(int i=0;i<32;i++){ m<<=1; m = m|(n & 1); n>>=1; } return m; } }; int main() { uint32_t n = 1; Solution sol; cout<<sol.reverseBits(n)<<endl; return 0; }
或:
class Solution { public: uint32_t reverseBits(uint32_t n) { uint32_t m = 0; for (int i = 0; i< 32 ; i++,n/=2) m = (m<<1) + (n%2); return m; } };
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