Leetcode 650 Python
2018-02-13 18:32
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解题思路:动态规划,dp[0],dp[1] = 0, 大于2时,dp[i]=min(dp[i],dp[j]+i/j) (i%j==0)
class Solution(object): def minSteps(self, n): """ :type n: int :rtype: int """ dp = [2**32-1]*(n+1) dp[0] = 0 dp[1] = 0 for i in range(2,n+1): for j in range(1,i): if i%j==0: dp[i] = min(dp[i],dp[j]+i/j) return dp
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