[leetcode] 473. Matchsticks to Square

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探讨如何判断一组不同长度的火柴是否能拼成一个正方形，介绍了一种使用递归深度优先搜索的方法，并通过优化提高了算法效率。

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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.

这道题是判断所有火柴能否围成一个正方形，题目难度为Medium。

所有火柴总长度必须是4的倍数，否则不能构成正方形。每根火柴必须用到，这里通过递归深度优先的方法逐个遍历所有火柴，依次将其加入到四条边上之后进行下一层遍历，如果最终四条边长度都达到边长长度则返回true，否则遍历完所有情况后返回false。可惜提交之后超时了，有什么方法能够优化呢？整个数组是无序的，如果我们优先遍历长度较长的火柴，这样包含这些火柴的情况如果不能构成正方形则会在较小的递归层数时就退出了，和无序的情况相比在递归时少了很多递归的可能，因此效率上有所提高，排序之后通过了。具体代码：

class Solution {
    bool chkSquare(const vector<int>& nums, vector<int>& len, int idx, int sdLen) {
        if(idx == nums.size()) {
            if(len[0] == sdLen && len[1] == sdLen && len[2] == sdLen)
                return true;
            else return false;
        }
        for(int i=0; i<4; ++i) {
            if(len[i] + nums[idx] <= sdLen) {
                len[i] += nums[idx];
                if(chkSquare(nums, len, idx+1, sdLen)) return true;
                len[i] -= nums[idx];
            }
        }
        return false;
    }
public:
    bool makesquare(vector<int>& nums) {
        if(nums.size() < 4) return false;
        vector<int> len(4, 0);
        int total = accumulate(nums.begin(), nums.end(), 0);
        if(total%4) return false;
        sort(nums.begin(), nums.end(), greater<int>());
        return chkSquare(nums, len, 0, total/4);
    }
};