LeetCode 279 Perfect Squares DP
Given an integer n
, return the least number of perfect square numbers that sum to n
.
A perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, 1
, 4
, 9
, and 16
are perfect squares while 3
and 11
are not
Solution
显然设 \(dp[i]\) 表示数字 i
的最少切分数,考虑转移的时候需要枚举此时的数字划分,由于其中一个数一定是平方数,所以只需要 \(O(\sqrt{n})\) 的复杂度即可
点击查看代码
class Solution { private: int dp[10005]; bool check(int x){ if(((int)sqrt(x)*(int)sqrt(x))==x)return true; return false; } public: int numSquares(int n) { if(n==1)return 1; dp[1] = 1; for(int i=2;i<=n;i++){ dp[i] = 9999999; if(check(i))dp[i] = 1; else{ for(int j=1;j<=sqrt(i);j++)dp[i] = min(dp[i], dp[j*j]+dp[i-j*j]); } } return dp[n]; } };