Perfect Squares

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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

解析：

递归求解每个状态，开始直接递归求解，因为有很多重复计算，超时，后来开辟一个dp数组记录每个状态的值减少冗余计算。

代码：

class Solution {public:    int numSquares(int n) {        int ans=INT_MAX;                vector<int>dp(n+1,0);        dp[1]=1;        return Numsq(n,dp);            }        int Numsq(int n,vector<int>&dp)    {        int ans=INT_MAX;        int temp=sqrt(n);        if (temp*temp==n)        return 1;        for (int i=temp; i>0; i--)        {            int tempans=0;            if (dp[n-i*i])            {                tempans=dp[n-i*i];            }            else            {                dp[n-i*i]=Numsq(n-i*i,dp);            }                        ans=min(ans,dp[n-i*i]+1);        }        dp[n]=ans;        return ans;    }    };