Perfect Squares
題意：
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.
解題思路：
遞迴法，複雜度相當高
public class Solution {
public int numSquares(int n) {
int res = n;
int num = 2;
while(num * num <= n) {
int a = n / (num * num);
int b = n % (num * num);
res = Math.min(res, a + numSquares(b));
num++;
}
return res;
}
}
DP法：
DP[i] 表示能形成 i 的最少perfect squares，遞迴式如下：
dp[i + j * j] = Math.min(dp[i + j * j], dp[i] + 1)
dp[i] + 1 表示只需要最少算出 i 的完美平方數，再加上j*j即可。
public class Solution {
public int numSquares(int n) {
if (n < 2) {
return 1;
}
int[] dp = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
for (int i = 0; i * i <= n; i++) {
dp[i * i] = 1;
}
for (int i = 0; i <= n; i++) {
for (int j = 1; i + j * j <= n; j++) {
// dp[i] + 1 表示只需要最少算出i的數，再加上j*j即可。
dp[i + j * j] = Math.min(dp[i + j * j], dp[i] + 1);
}
}
return dp[n];
}
}
數論法：
This problem is completely solved, in the sense of being reduced to simple checks of a number's prime factorization. 一個自然數可能是:
一個平方數 if and only if each prime factor occurs to an even power in the number's prime factorization.
兩個平方數之和 if and only if each prime factor that's 3 modulo 4 occurs to an even power in the number's prime factorization.
三個平方數之和 if and only if it's not of the form 4a(8b+7) with integers a and b. 四個平方數之和. Period. No condition. You never need more than four.