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### leetcode Perfect Squares

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

dp,设dp[i]为和i的最少完全平方数。

dp[i] = min(dp[i - j*j] + 1) {j *j <i}

### code

C++

Python同样的代码会超时。。。