leetcodePerfect Squares (#279)

Posted 2020-08-23 Dragonir

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.

 

解析：

利用动态规划解决此问题：对于要求的当前节点而言都是从前面的节点转移过来的，只是这些转移节点并非一个，而是多个，比如1*1，2*2，3*3，，，那么相应的res[i-1]、res[i-4]、res[i-9]等等都是转移点。从这些候选项中找到最小的那个，然后加1即可。

 

算法实现代码：

//#include "stdafx.h"
#include <iostream>
#include <cstdio>
#include <climits>
#include <ctime>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <cstdlib>
#include <windows.h>
#include <string>
#include <cstring>
#include <cmath>

using namespace std;

class Solution {
public:
    int numSquares(int n) {
        vector<int> res(n + 1);
        for (int i = 0; i <= n; ++i){
            res[i] = i;
            for (int j = 1; j * j <= i; ++j){
                res[i] = min(res[i - j * j] + 1, res[i]);
            }
        }
        return res[n];
    }
};

int main(){
	Solution s;
	cout<<s.numSquares(13);
	return 0;
}