Leetcode 338. Counting Bits

题目链接

问题描述

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].

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

假设返回的序列是dp，通过观察0到num序列可以发现，如果这个数是奇数，则dp[i] = dp[i-1] + 1；如果这个数是偶数则分为①可以被4除尽，②不可以被4除尽两种情况。

可以被4除尽除尽时，dp[i] = dp[i-1] - X + 1；其中X是这个数可以被2除尽的次数

不可以被4除尽时，dp[i] = dp[i-1]

代码如下

class Solution {
public:
	vector<int> countBits(int num) {
		vector<int> dp;
		for (int i = 0; i <= num; i++) {
			dp.push_back(0);
		}
		if (num == 0) return dp;
		for (int i = 1; i <= num; i++) {
			if (i % 2 != 0) dp[i] = dp[i - 1] + 1;  // i 为奇数
			else if (i % 4 != 0) dp[i] = dp[i - 1];  // i为不能整除4的偶数
			else {     // i 为能整除4的偶数
				int twos = 0;
				int tmp = i;
				while (tmp % 2 == 0) {
					twos++;
					tmp /= 2;
				}
				dp[i] = dp[i - 1] - twos + 1;
			}
		}
		return dp;
	}
};