LeetCode 279. Perfect Squares

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.
分析：建立一个n+1长度的数组dp，dp[i]表示i这个数构成平方和需要数字的最小个数。
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
当j*j=i的时候，dp[i]的值就直接为1
从2一直到n可以计算出dp[i]的所有值。。
最后return dp[n]的值。

class Solution {
public:int numSquares(int n) {vector<int> dp(n+1);dp[1] = 1;for(int i = 2; i <= n; i++) {int temp = 99999999;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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