leetcode014-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:
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.

解题思路：动态规划，dp[i]代表数字i的二进制含有多少个1

  如果i是奇数的话，i-1是偶数，那个i-1的二进制的最低位一定是0，加1并不会进位，所以dp[i] = dp[i-1]+1

  如果i是偶数的话，i会与i/2的二进制拥有相同个数的1，因为i/2左移一位，最后补0，就是i，所以dp[i] = dp[i/2]

class Solution {public:    vector<int> countBits(int num) {        vector<int> dp(num+1,0);        for (int i = 1; i <= num; i++) {            if (i%2==0) {                dp[i] = dp[i/2];            } else {                dp[i] = dp[i-1]+1;            }        }        return dp;    }};