leetcode--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.

思路：建立一个长度为n+1的数组，i从0循环到n，j从1循环到i+j*j <= n的位置，然后每次更新数组中下标为i+j*j的值,数组中下标为i的元素代表正整数i能被多少个完全平方数组成。

class Solution {
public:
int numSquares(int n) {
vector<int> dp(n + 1, INT_MAX);
dp[0] = 0;
for (int i = 0; i <= n; ++i) {
for (int j = 1; i + j * j <= n; ++j) {
dp[i + j * j] = min(dp[i + j * j], dp[i] + 1);
}
}
return dp.back();
}
};
思路：递归，遍历所有比n小的完全平方数，然后对n与完全平方数的差值递归调用函数，目不断更新最终结果，最后找到最小的那个，效率很慢，但最容易实现
class Solution {
public:
int numSquares(int n) {
int res = n, num = 2;
while (num * num <= n) {
int a = n / (num * num), b = n % (num * num);
res = min(res, a + numSquares(b));
++num;
}
return res;
}
};