版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/Wilder_ting/article/details/86180255
leetcode338—Counting Bits
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 1:
Input: 2 Output: [0,1,1]
Example 2:
Input: 5
Output: [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.
想法:统计0-num之间的二进制表示中1的个数。用result[i]表示i的二进制表示中1的个数。利用动态规划,找出状态转移表达式result[i]=result[i&i-1]+1
class Solution { public: vector<int> countBits(int num) { vector<int> result(num+1); result[0] = 0; for(int i = 1 ;i <= num ; i++){ result[i] = result[i & i-1] + 1; } return result; } };