[LeetCode]Perfect Squares

题目链接：Perfect Squares

题目内容：

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.

题目解法：

最初看到这个题，我想到的是回溯法，剪了半天的枝还是各种超时，后来参考了xudli的解法，才知道这道题用动态规划更合适。

我们设d[i]=a表示数字i对应的least number of perfect square numbers为a，则显然d[1]=1为初始条件，接着我们用背包问题的思想，对于i从2到n，尝试向其中放入perfect square number j，j从1开始枚举，每次放入后得到的结果为d[i - j*j]+1，也就是在不放入j的least number基础上+1得到d[i]的最小值，对于不同的j，我们应该选取其中最小的那个，也就是说：d[i] = min{d[i-j*j],j=1,2,3...,j*j<=i}。

最后，d[n]就是结果。

代码如下：

class Solution {
public:int getMin(int a, int b){return a < b ? a : b;}int numSquares(int n) {int *d = new int[n+1];d[1] = 1; // d[i]表示数字i的Prefect Seuares值。for(int i = 2; i <= n; i++){int j = 1;int min = 99999999;while(j*j <= i){if(j*j == i){min = 1;break;}min = getMin(min,d[i-j*j] + 1);j++;}d[i] = min;}return d[n];}
};

转载于:https://www.cnblogs.com/aiwz/p/6154018.html

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