LeetCode No.279 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.

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题目链接：https://leetcode.com/problems/perfect-squares/

题目大意：求n的最少平方和。

思路：动态规划。dp[i] 表示i的最少平方和。

初始化dp数组：

vector <int> dp ( n + 1 , 0 ) ;
vector <int> v ;
for ( int i = 1 ; i <= sqrt ( n ) ; i ++ )
{
    int p = i * i ;
    dp[p] = 1 ;
    v.push_back ( p ) ;
}

动态规划：

int sz = v.size() ;
for ( int i = 2 ; i <= n ; i ++ )
{
    int ans = INT_MAX ;
    for ( int j = 0 ; j < sz ; j ++ )
    {
        if ( v[j] > i )
            break ;
        ans = min ( ans , 1 + dp[i-v[j]] ) ;
    }
    dp[i] = ans ;
}

参考代码：

class Solution {
public:
    int numSquares(int n) {
        vector <int> dp ( n + 1 , 0 ) ;
        vector <int> v ;
        for ( int i = 1 ; i <= sqrt ( n ) ; i ++ )
        {
            int p = i * i ;
            dp[p] = 1 ;
            v.push_back ( p ) ;
        }
        int sz = v.size() ;
        for ( int i = 2 ; i <= n ; i ++ )
        {
            int ans = INT_MAX ;
            for ( int j = 0 ; j < sz ; j ++ )
            {
                if ( v[j] > i )
                    break ;
                ans = min ( ans , 1 + dp[i-v[j]] ) ;
            }
            dp[i] = ans ;
        }
        return dp[n] ;
    }
};