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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).
简单的做法就是遍历一遍，但是这个方法很低效，貌似也没有什么更好的办法了
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class Solution {
public:
uint32_t reverseBits(uint32_t n) {
if(0 == n)
{
return 0;
}
int res = 0;
for(int i = 0; i < 32; i++)
{
if( n & 1)
{
res += (1 << (31-i));
}
n = n >> 1;
}
return res;
}
};
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Number of 1 Bits
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Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight).
For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3.
这里注意 n&(n-1)可以将最后一个1消掉，所以不用全部遍历一遍，有多少个1，就循环多少次
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class Solution {
public:
int hammingWeight(uint32_t n) {
if(0 == n)
{
return 0;
}
int res = 0;
while(n)
{
n = n&(n-1);
res++;
}
return res;
}
};
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Reverse Integer
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Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321 这里计算过程并不复杂，但是注意是否会超过范围
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class Solution {
public:
int reverse(int x) {
long long xx = x; //防止负数超过范文
if(0 == x)
{
return 0;
}
int flag = 0;
if(x < 0)
{
flag = 1;
xx = -x;
}
stack
temp;
while(xx)
{
temp.push(xx%10);
xx = xx/10;
}
long long index = 1;
long long res = 0;
while(!temp.empty())
{
res += temp.top()*index;
index = index*10;
temp.pop();
}
if(flag)
{
res = res*(-1);
if(res < INT_MIN)
return 0;
}
if(res > INT_MAX)
{
return 0;
}
return res;
}
};
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