Problem

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:

1. The length sum of the given matchsticks is in the range of0to10^9.
2. The length of the given matchstick array will not exceed15

思路

• 一开始思路没想出来， 当时想的是如何通过index一条边一条边解决， 很混乱
• 还是要寻求解决方案入手
  • 每个dfs路径都是一个矩形的组合，
  • 每个组合4条边
  • 每次到数组最后，检测当前边的组合是否符合正方形
• 优化
  • 对数组排序
  • index从数组最后开始， 如果其中一条边大于需要符合标准的边长。
  • 这样能从一开始就过滤掉不符合的。

    public boolean makesquare(int[] nums) {
        if (nums == null || nums.length < 4) return false;

        Arrays.sort(nums);

        int sum = 0;
        for (int num: nums) sum += num;
        if (sum % 4 != 0) return false;
        return helper(nums, nums.length - 1, 0, 0, 0, 0, sum / 4);
    }

    private boolean helper(int[] nums, int index, int e1, int e2, int e3, int e4, int width) {
        if (e1 > width || e2 > width || e3 > width || e4 > width) return false;

        if (index == -1 && e1 == width && e2 == width && e3 == width && e4 == width) return true;

        return helper(nums, index - 1, e1 + nums[index], e2, e3, e4, width) ||
                helper(nums, index - 1, e1, e2 + nums[index], e3, e4, width) ||
                helper(nums, index - 1, e1, e2, e3 + nums[index], e4, width) ||
                helper(nums, index - 1, e1, e2, e3, e4 + nums[index], width);
    }