leetcode 279. Perfect Squares

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279. Perfect Squares

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Total Accepted: 33524 Total Submissions: 102649 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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题解：

转自：http://www.lai18.com/content/1370786.html

 

四平方和定理这种可遇不可求的做法就不说了，说说dp:

 

 

dp[i]代表组成i的最少平方数的个数。至于状态转移，我们可以这样想：组成这个数的最小值要么就是本身，要么就是前面某一个数+一个平方数（所以看作值加上1）,类似0-1背包的思想。

 

Submission Details

600 / 600 test cases passed.	Status:  Accepted
Runtime: 461 ms

class Solution {
public:
    int numSquares(int n) {
        vector<int> dp(n+1,INT_MAX);
        dp[0] = 0;  
        dp[1] = 1;  
        for(int i=2; i<=n; i++) {  
            for(int j = 1;j * j <= i ;j++){
                dp[i] = min(dp[i], dp[i - j * j] + 1);  
            }  
        }  
        return dp[n];  
    }
};