Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...
) which sum to n.
Example 1:
Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.
Example 2:
Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.
方法1:动态规划
int[] dp = new int[n+1];
dp[0] = 0;
for (int i = 1; i <= n; i++) {
int min = Integer.MAX_VALUE;
for (int k = 1; k * k <= i; k++) {
min = Math.min(min, 1+dp[i-k*k]);
}
dp[i] = min;
}
return dp[n];
时间复杂度:O(n^2)
空间复杂度:O(n)