Matchsticks to Square && Grammar: reverse an primative array

Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.

Example 1:
Input: [1,1,2,2,2]
Output: true

Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.
Example 2:
Input: [3,3,3,3,4]
Output: false

Explanation: You cannot find a way to form a square with all the matchsticks.
Note:
The length sum of the given matchsticks is in the range of 0 to 10^9.
The length of the given matchstick array will not exceed 15.

DFS, my solution is to fill each edge of the square one by one. DFS to construct the 1st, then 2nd, then 3rd, then 4th. For each edge I scan all the matches but only pick some, the others are discarded. These brings about two disadvantages: 1. I need to keep track of visited elem. 2. after the 1st edge is constructed, I have to re-scan to make the 2nd, 3rd, 4th, thus waste time! So my code get TLE in big case

public class Solution {
    public boolean makesquare(int[] nums) {
        if (nums==null || nums.length==0) return false;
        int sideLen = 0;
        for (int num : nums) {
            sideLen += num;
        }
        if (sideLen % 4 != 0) return false;
        sideLen /= 4;
        return find(nums, sideLen, 0, 0, new HashSet<Integer>());
    }
    
    public boolean find(int[] nums, int sideLen, int completed, int curLen, HashSet<Integer> visited) {
        if (curLen > sideLen) return false;
        if (curLen == sideLen) {
            completed++;
            curLen = 0;
            if (completed==4 && visited.size()==nums.length) return true;
            if (completed==4 && visited.size()<nums.length) return false;
        }
        for (int i=0; i<nums.length; i++) {
            if (!visited.contains(i)) {
                visited.add(i);
                if (find(nums, sideLen, completed, curLen+nums[i], visited))
                    return true;
                visited.remove(i);
            }
        }
        return false;
    }
}

　　

Better Solution: DFS, but only scan all the matches once! If a match can not make the 1st edge, then we do not discard it, we try 2nd, 3rd, 4th edge

referred to: https://discuss.leetcode.com/topic/72107/java-dfs-solution-with-explanation

public class Solution {
    public boolean makesquare(int[] nums) {
        if (nums == null || nums.length < 4) return false;
        int sum = 0;
        for (int num : nums) sum += num;
        if (sum % 4 != 0) return false;
        
        return dfs(nums, new int[4], 0, sum / 4);
    }
    
    private boolean dfs(int[] nums, int[] sums, int index, int target) {
        if (index == nums.length) {
            if (sums[0] == target && sums[1] == target && sums[2] == target) {
            return true;
            }
            return false;
        }
        
        for (int i = 0; i < 4; i++) {
            if (sums[i] + nums[index] > target) continue;
            sums[i] += nums[index];
            if (dfs(nums, sums, index + 1, target)) return true;
            sums[i] -= nums[index];
        }
        
        return false;
    }
}