Leetcode Perfect Squares-优快云博客

探讨如何通过递归和动态规划方法找到给定整数的最小完美平方数之和，涉及数学算法与编程实现。

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Problem

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.

Solution 1 - pure recursion

求n 的perfect square, 可以化简成求 n 减去若干个 i*i之后的数的perfect square.

class Solution {
public:
    int numSquares(int n) {
        if(n <= 1) return n;
        
        int rst = n;
        for( int i = 2; i*i <= n; i++){
            int a = n / (i * i), b = n % (i * i);
            rst = min (rst, numSquares(b) + a);
        }
        return rst;
    }
};

Solution 2 - recursion with memory

用一个数字，纪录上面的过程中重复值。

class Solution {
    int helper( int n, vector<int>& arr){
        if(arr[n] != -1) return arr[n];
        
        int rst = n;
        for( int i = 2; i*i <= n; i++){
            int a = n / (i * i), b = n % (i * i);
            rst = min (rst, helper(b, arr) + a);
        }
        arr[n] = rst;
        return rst;
               
    }
public:
    int numSquares(int n) {
        if(n <= 1) return n;
        vector<int> arr(n+1, -1);
        helper(n, arr);
        return arr[n];
    }
};

Solution 3 - DP

其实上面的解法二已经很接近DP了

dp[ x + y*y ] = min( dp[ x+ y*y ] , dp[ x ] + 1 );

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