279. Perfect Squares
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参考： https://www.geeksforgeeks.org/minimum-number-of-squares-whose-sum-equals-to-given-number-n/
先写出递归解 O(n!) complexity
再用一个数组存储中间参数，然后获得DP解
public class PerfectSquares {
/*
public int numSquares(int n) {
if (n == 0) return 0;
int res = n; // 1*1 + 1*1 +... + 1*1
for (int i = 1; i*i <= n; i++) {
if (i*i > n) break;
else {
res = Math.min(res, 1 + numSquares(n-i*i));
}
}
return res;
}
*/
public int numSquares(int n) {
// dp[k] == the Minimum number of squares whose sum equals to given number k
if (n == 0) return 0;
else if (n == 1) return 1;
else if (n == 2) return 2;
else if (n == 3) return 3;
int[] dp = new int[n+1]; // 0, 1, 2, ..., n
dp[0] = 0;
dp[1] = 1;
dp[2] = 2;
dp[3] = 3;
// compute dp[i]
for (int i = 4; i <= n; i++) {
dp[i] = i;
for (int x = 1; x * x <= i; x++) {
dp[i] = Math.min(dp[i], 1 + dp[i - x * x]);
}
}
return dp[n];
}
}
解析： 前者是recursive solution。后者用了递归。
