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
.
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
public class Solution {
public int numSquares(int n) {
int dp[] = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
for (int i = 0; i < n; i++) {
for (int j = 1; i + j * j <= n; j++) {
dp[i + j * j] = Math.min(dp[i + j * j], 1 + dp[i]);
}
}
return dp[n];
}
}
============
public class Solution {
public int numSquares(int n) {
int m = n;
while( m % 4 == 0 )
m = m>>2;
if(m % 8 == 7)
return 4;
int sqrtOfn = (int) Math.sqrt(n);
if(sqrtOfn * sqrtOfn == n)//Is it a Perfect square?
return 1;
else{
for(int i = 1; i <= sqrtOfn; ++i){
int remainder = n - i*i;
int sqrtOfNum = (int) Math.sqrt(remainder);
if(sqrtOfNum * sqrtOfNum == remainder)
return 2;
}
}
return 3;
}
}
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