problem:https://leetcode.com/problems/counting-bits
爬台阶类型。我的做法是num拆分为比它小的最大2^n的值加上另一个数，然后这两个的dp叠加。
class Solution { public: vector<int> countBits(int num) { vector<int> dp(num + 1); dp[0] = 0; int cur = 1; int next = 1; int idx = 0; for(int i = 1;i <= num;i++) { if(i == next) { dp[i] = 1; idx++; cur = next; next = pow(2, idx); } else { dp[i] = dp[i - cur] + 1; } } return dp; } };
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原文：https://www.cnblogs.com/fish1996/p/11332352.html