[WXM] LeetCode 473. Matchsticks to Square

最新推荐文章于 2018-09-06 16:15:14 发布

WX_ming

最新推荐文章于 2018-09-06 16:15:14 发布

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本文链接：https://blog.youkuaiyun.com/WX_ming/article/details/81292143

本文探讨了如何通过算法判断给定长度的火柴棍能否拼接成一个正方形。利用深度优先搜索策略，对输入数据进行排序并分配到四个组中，确保每组长度相等从而构成正方形。

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473. Matchsticks to Square

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.

Approach

题目大意：给你一组数据，问你是否能够拼接成一个正方形
思路方法：首先将数据从大到小排序，开辟四个槽，然后将一个数据放入其中一个槽内，再DFS判断此时是否能够使得四个槽的数据相同，不相同则将槽内的这个数据移除

Code

class Solution {
public:
    bool makesquare(vector<int>& nums) {
        if (nums.size() < 4)return false;
        int sum = accumulate(nums.begin(),nums.end(),0);
        if (sum % 4||sum<=0)return false;
        vector<int>sums(4, 0);
        sort(nums.rbegin(), nums.rend());
        return helper(nums, sums, 0, sum / 4);
    }
    bool helper(vector<int>&nums,vector<int>&sums,int pos,int target) {
        if (pos >= nums.size()) {
            return sums[0] == target&&sums[1] == target&&sums[2] == target;
        }
        for (int i = 0; i < 4; i++) {
            if (sums[i] + nums[pos] > target)continue;
                sums[i] += nums[pos];
            if (!helper(nums, sums, pos + 1, target)) //递归判断是否可行
                sums[i] -=nums[pos];//不行则将其移除
            else return true;
        }
        return false;
    }
};