【LeetCode 473】 Matchsticks to Square

題目描述

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.

思路

dfs，卡在剪枝，一直超時。
對對數組從大到小排序會減少時間。
第一種方法裏，如果cur==0時 最大的數字找不到安排，可以直接返回false了。

代碼

方法一：

class Solution {
public:
    bool makesquare(vector<int>& nums) {
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if (sum % 4) return false;
        int target = sum / 4;
        int n = nums.size();
        vector<int> vis(n, 0);
        sort(nums.begin(), nums.end(), greater<int>());
        return dfs(nums, target, vis, 0, 0);
    }
    
    bool dfs(vector<int>& nums, int target, vector<int>& vis, int cur, int cnt) {
        if (cur == target) {
            cnt++;
            cur = 0;
        }
        if (cnt == 3) return true;
        for (int i=0; i<nums.size(); ++i) {
            if (vis[i]) continue;
            if (cur + nums[i] > target) continue;
            vis[i] = 1;
            if (dfs(nums, target, vis, cur+nums[i], cnt)) return true;
            vis[i] = 0;
            if (cur == 0) return false;
        }
        
        return false;
    }
};

方法二：

class Solution {
public:
    bool makesquare(vector<int>& nums) {
        if (nums.empty()) return false;
        int n = nums.size();
        int sum = accumulate(nums.begin(), nums.end(), 0);
        if (sum % 4) return false;
        vector<int> sums(4, 0);
        sort(nums.begin(), nums.end(), greater<int>());
        return dfs(nums, sums, 0, n, sum/4);
    }
    
    bool dfs(vector<int>& nums, vector<int>& sums, int cnt, int n, int target) {
        if (cnt == n && sums[0] == target && sums[1] == target && sums[2] == target && sums[3] == target) {
            return true;
        }else if (cnt == n) {
            return false;
        }
        
        for (int i=0; i<4; ++i) {
            if (sums[i] + nums[cnt] > target) continue;
            sums[i] += nums[cnt];
            if (dfs(nums, sums, cnt+1, n, target)) return true;
            sums[i] -= nums[cnt];
        }
        
        return false;
    }
};