(DP 线性DP) leetcode 338. 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.
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这个大佬已经解释清楚了,找找规律就行。
/** * 题目已经强调了需要O(n)的复杂度,只能遍历一遍,可以考虑动态规划 * 根据题目意思,先手动画一下数字和2进制的具体映射关系 * 数字 0 1 2 3 4 5 6 7 8 * 二进 0 1 10 11 100 101 110 111 1000 * 1个数 0 1 1 2 1 2 2 3 1 * 根据递推效果,看着好像没有什么规律 * 但是仔细思考下,10进制转2进制必须要除以2,有些能整除,有些不能整除 * 不能整除的3的1个数=3/1=数字1的1个数+1 * 能整除的4的的1个数=4/2=数字2的1个数 * 拿其他数字验证后发现的确是这个规律,得到动态规划状态转移方程: * int d = i / 2; * int m = i % 2; * if (m == 0) { * dp[i] = dp[d]; * } else { * dp[i] = dp[d] + 1; * } * * @param num * @return */
C++代码:
class Solution { public: vector<int> countBits(int num) { vector<int> dp(num+1,0); for(int i = 1; i <= num; i++){ int d = i / 2; int m = i % 2; if(m == 0) dp[i] = dp[d]; else dp[i] = dp[d] + 1; } return dp; } };