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.

• n = a + b * b
• a：n之前的某个值
• n必定由一个值+完美平方数组成
• 这个值为子问题，也是最佳问题

int numSquares(int n) {
    //开辟n个大小的数组，初始化为0
    int *result = calloc((n + 1), sizeof(int));
    //填入平方的值
    int sqrtn = sqrt(n);
    result[1] = 1;
    for (int i = 2; i <= n; i++){
        int min = n;
        int j = 1;
        while(j * j <= i){
            if (min > result[i - j*j] + 1)
                min = result[i - j*j] + 1;
            j++;
        }
        result[i] = min;
    }
    return result[n];
}