
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, …) which sum to n.
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| Example 1:
Input: n = 12
Output: 3
Explanation: 12 = 4 + 4 + 4.
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| Example 2:
Input: n = 13
Output: 2
Explanation: 13 = 4 + 9.
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Solution#
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| class Solution {
public int numSquares(int n) {
int[] dp = new int[n + 1];
dp[0] = 0;
dp[1] = 1;
for (int i = 2; i < dp.length; i++) {
dp[i] = Integer.MAX_VALUE;
for (int j = 1; j * j <= i; j++) {
dp[i] = Math.min(dp[i], dp[i - j * j] + 1);
}
}
return dp[n];
}
}
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Solution 2021-11-30#
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| class Solution {
public int numSquares(int n) {
int[] dp = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
int max_square_index = (int) Math.sqrt(n) + 1;
int[] squares = new int[max_square_index];
for (int i = 1; i < squares.length; i++) {
squares[i] = i * i;
}
for (int i = 1; i < dp.length; i++) {
for (int j = max_square_index - 1; j > 0; j--) {
if (i - squares[j] >= 0) {
int at = i - squares[j];
dp[i] = Math.min(dp[i], dp[at] + 1);
}
}
}
return dp[n];
}
}
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