leetcode -- 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.

Example 1:

Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.
Example 2:

Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/perfect-squares
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路：

1. dp[i]表示构成i这个数构成平方和需要数字的最小个数

2. 当j*j<i的时候，temp中保存j从1开始每加1得到的dp[i-j*j] + 1的最小值

3. 当j*j=i的时候，dp[i]的值就直接为1

class Solution {
public:int numSquares(int n) {vector<int> dp(n+1);dp[1] = 1;for(int i = 2; i <= n; i++){int temp = 999999999;for(int j = 1; j*j <= i; j++){if(j*j == i){temp = 1;break;}temp = min(temp, dp[i-j*j]+1);}dp[i] = temp;}return dp[n];}
};

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