[Leetcode]Perfect Squares(DP and Math Solution)
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**Perfect Squares My Submissions Question
Total Accepted: 16355 Total Submissions: 55778 Difficulty: Medium
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.
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题意很简单,就是找出一个正整数,找出最少用几个完全平方数(不包括0)可以表示它,用动态规划做就可以了,然而用时788ms。
其实最小可以缩到4ms,根据[Lagrange’s four-square theorem]
任一个自然数都可以用4个完全平方数表示出来,所以这里分情况讨论就好了,注意可能包括0!
下面贴出788ms的代码,以及4ms的代码。
class Solution {public: int numSquares(int n) { vector<int> dp(n + 1,0); for(int i = 1;i <= n;++i){ dp[i] = i; } for(int i = 1;i <= n;++i){ int m = sqrt(i); for(int j = m;j >= 1;--j){ dp[i] = min(dp[i],dp[i - pow(j,2)] + 1); } } return dp[n]; }};
Legendre’s three-square theorem
class Solution { private: int is_square(int n) { int sqrt_n = (int)(sqrt(n)); return (sqrt_n*sqrt_n == n); }public: // Based on Lagrange's Four Square theorem, there // are only 4 possible results: 1, 2, 3, 4. //根据Lagrange's Four Square theorem,有且只 //有4种可能的结果:1,2,3,4 int numSquares(int n) { // If n is a perfect square, return 1. //如果n是个完全平方数,返回1 if(is_square(n)) { return 1; } // The result is 4 if n can be written in the // form of 4^k*(8*m + 7). Please refer to //如果n可以被写成4^k*(8*m + 7)的形式, //那么结果就是4,请参考Legendre's three-square theorem //这里多说一句,Legendre's three-square theorem是说 //如果n不可以被写成4^k*(8*m + 7)的形式,那么n就可以 //用三个整数的平方和表示。 //这里判断的是:如果n可以被写成4^k*(8*m + 7)的形式 //那么就直接返回4。这里的差别主要是要考虑Legendre's //three-square theorem 里说的是整数,可能包括0,比如说 //对于5 = 2^2 + 1^2 + 0^2.而我们要求的则不包括0. //感觉这里需要认真体会下这个细节:如果n可以被写成 //4^k*(8*m + 7)的形式,那么就直接返回4 while ((n & 3) == 0) // n%4 == 0 { n >>= 2; //如果n能被4整除,那么n和n/4, //其返回的值是相同的! } if ((n & 7) == 7) // n%8 == 7 { return 4; } // Check whether 2 is the result. //检查2是不是结果 int sqrt_n = (int)(sqrt(n)); for(int i = 1; i <= sqrt_n; i++) { if (is_square(n - i*i)) { return 2; } } return 3; //其他情况返回3 } };
除此之外,这道题还有另外两种解法,Static Dynamic Programming和BFS。这里就不贴出来了,有兴趣的可以戳这里
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