选坐标进行gcd运算 1799. Maximize Score After N Operations

You are given nums, an array of positive integers of size 2 * n. You must perform n operations on this array.

In the ith operation (1-indexed), you will:

• Choose two elements, x and y.
• Receive a score of i * gcd(x, y).
• Remove x and y from nums.

Return the maximum score you can receive after performing n operations.

The function gcd(x, y) is the greatest common divisor of x and y.

 

Example 1:

Input: nums = [1,2]
Output: 1
Explanation: The optimal choice of operations is:
(1 * gcd(1, 2)) = 1

Example 2:

Input: nums = [3,4,6,8]
Output: 11
Explanation: The optimal choice of operations is:
(1 * gcd(3, 6)) + (2 * gcd(4, 8)) = 3 + 8 = 11

Example 3:

Input: nums = [1,2,3,4,5,6]
Output: 14
Explanation: The optimal choice of operations is:
(1 * gcd(1, 5)) + (2 * gcd(2, 4)) + (3 * gcd(3, 6)) = 1 + 4 + 9 = 14

忘了怎么求全部方案中的最大值，囧
先用回溯法求出所有解，然后每次都保留max，从而取最大值。有道理啊

//https://leetcode.com/problems/maximize-score-after-n-operations/discuss/1200095/Java-Backtrack-with-memoization
class Solution {
    int res;
    int n;
    HashMap<String, Integer> memo;
    public int maxScore(int[] nums) {
        res = 0;
        n = nums.length/2;
        memo = new HashMap<>();
        return backtrack(nums, 1, new HashSet<>());
        //return res;
    }
    
    private int backtrack(int [] nums, int i, HashSet<Integer> set) {
        if(i > n) {
            return 0;
        }
        
        String key = convert(i,set);
        System.out.println("key是i加上set = " + key);
        //就是先判断一下有没有进行到过这一步，有的话就取出来
        if(memo.containsKey(key)) {
             return memo.get(key);
        }
           
        
        int val=0, res=0;
        for(int j=0; j<nums.length; j++) { 
            if(!set.contains(j)) {
                for(int k=j+1; k<nums.length; k++) {
                    val=0;
                    if(!set.contains(k)) {
                        set.add(j); set.add(k);
                        val = i * (gcd(nums[j], nums[k])) + backtrack(nums, i+1, set);
                        //每次都只保留最大值
                        res = Math.max(res, val);
                        set.remove(j); set.remove(k);
                    }  
                }
            }
        }
        
        memo.put(key, res);
        return res;
    }
    
    private int gcd(int a, int b) {
        
        return (a%b)==0 ? b : gcd(b, a%b);
    }
    
    private String convert(int i, HashSet<Integer> set) {
        return Integer.toString(i) + set.toString();
    }
}