leetcode:Counting Bits
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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].
Space complexity should be O(n).
Example:
For num = 5 you should return [0,1,1,2,1,2].
Follow up:
It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?Space complexity should be O(n).
Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.
给定num,对于每个 0<= i <=num , 计算i的2进制中1的个数。
若对每个数,进行每位的判断,时间复杂度是O(n*logn)。
可以从每个数的最低位开始分析,观察到
1
10
11
100
101
110
111
1000
例如1001001 ,它的二进制1的个数等于100100 总二进制个数 + 1 。即循环地在每个数后面加0、1可得接下来的数字。
因此第i位就是第i/2位+(i%2)的值。
代码:
class Solution {
public:
vector<int> countBits(int num) {
vector<int>vec(num+1,0);
vec[0]=0;
for(int i = 1; i <= num; i++){
vec[i]=vec[i/2]+(i % 2);
}
return vec;
}
};
public:
vector<int> countBits(int num) {
vector<int>vec(num+1,0);
vec[0]=0;
for(int i = 1; i <= num; i++){
vec[i]=vec[i/2]+(i % 2);
}
return vec;
}
};
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