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
.
计算一个数最少由多少个完全平方数相加而成。
简单DP,从1开始向上转移。
class Solution {
public:
int numSquares(int n) {
vector<int> dp(n + 1, INT_MAX);
dp[0] = 0;
for(int i = 1; i <= n; i++ )
for(int j = 1; j <= sqrt(i); j++)
dp[i] = min(dp[i], dp[i - j * j] + 1);
return dp[n];
}
};