《leetcode》:Perfect Squares
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题目
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.
思路
这个题自己也没有做出来,哎,水平不够也;
思路来自于:https://segmentfault.com/a/1190000003768736
public int numSquares(int n) { int[] dp=new int[n+1]; Arrays.fill(dp, Integer.MAX_VALUE); //将所有平方和的结果置为1 for(int i=0;i*i<=n;i++){ dp[i*i]=1; } /* * 如果一个数x可以表示为一个任意数a加上一个平方数bxb,也就是x=a+bxb,那么能组成这个数x最少的平方数个数, * 就是能组成a最少的平方数个数加上1(因为b*b已经是平方数了)。 * */ for(int a=0;a<n;a++){ for(int b=0;a+b*b<=n;b++){ dp[a+b*b]=Math.min(dp[a]+1, dp[a+b*b]); } } return dp[n]; }
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