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:这道题类似于coin changing,用动态规划来做,思路就是,dp[n]取值于最小的dp[n - i] (i为小于n的可换的硬币值,在这里就是小于n的平方数),因此取最小的时候要每个硬币值都要尝试,所以里面需要一个循环,外面需要一个由j : 1 -> n的求dp[j]的大循环,代码如下:
public int numSquares(int n) {
int[] dp = new int[n + 1];
if (n == 0) return 0;
dp[0] = 0;
for (int i = 1; i <= n; i++) {
int min = Integer.MAX_VALUE;
for (int j = 1; j * j <= i; j++) {
min = Math.min(min, dp[i - j * j] + 1);
}
dp[i] = min;
}
return dp[n];
}