题目描述:
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them as an array.
Example:
For num = 5 you should return [0,1,1,2,1,2].
分析:
考虑整数a,a-1;
当他们转换为2进制时,如果a从小到大是i位‘0’然后一个1,然后一串数字x,那么a-1必然是i个‘1’,一个‘0’,然后一串x,i大于等于0,包含了所有情况。当这两个数取与时,明显看出会比a的‘1’的总数小一,由此递推即可。
代码如下:
vector<int> countBits(int num) {
vector<int> res(num+1,0);
res[0]=0;
res[1]=1;
for(int i=2;i<=num;i++){
int temp=i&(i-1);
res[i]=res[temp]+1;
}
return res;
}