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.
solution:
similar to 0/1 knapsack problem, use dp to solve this.
public class Solution {
public int numSquares(int n) {
int[] dp = new int[n+1];
dp[1] = 1;
for(int i=2;i<=n;i++){
int min = Integer.MAX_VALUE;
int j = 1;
while(j*j<=i) {
if(j*j == i){
min = 1;
break;
}
min = Math.min(min, dp[i-j*j]+1);
++j;
}
dp[i] = min;
}
return dp[n];
}
}

本文探讨了如何找出构成给定正整数n所需的最少完全平方数的数量,并提供了一个使用动态规划解决该问题的Java实现。
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