[LeetCode 659] Split Array into Consecutive Subsequences

You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split.

Example 1:

Input: [1,2,3,3,4,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3
3, 4, 5

Example 2:

Input: [1,2,3,3,4,4,5,5]
Output: True
Explanation:
You can split them into two consecutive subsequences : 
1, 2, 3, 4, 5
3, 4, 5

Example 3:

Input: [1,2,3,4,4,5]
Output: False

Note:

1. The length of the input is in range of [1, 10000]

 

分析

这一题目还是挺新颖的，我第一时间也没有想到能够在时间复杂度内的解法。

先说一下最初想到的解法，我们使用二维数组记录所有的subsequences。遍历nums，每一个数nums[i]都在二维数组中寻找可以append的最短的数组，如果找不到，就以这个数为开头新增一个数组。

例如：1,2,3,3,4,5

[1]->[1,2]->[1,2,3]->[1,2,3] -> [1,2,3] -> [1,2,3]

                               [3]           [3,4]         [3,4,5]

这种解法会超时，因为每次我们都需要寻找最短的可用数组，最差条件下可能会有O(n*n)的复杂度，所以会超时。

 

在网上看到了其他人的解法，其实我们不必使用二维数组记录所有的subsquence，只需要使用map记录每一个subsquence最后一位数，即可。但是在每一次新起一个开头时，例如上个例子中的nums[4]，都需要直接判断之后是否有nums[4] + 1, nums[4] + 2存在。如果存在就直接将这三个数字组成一个新的数组。

 

Code

超时的解法

class Solution {
public:
    bool isPossible(vector& nums) {
        vector> subs;
        int length = nums.size();
        if (length < 3)
            return false;
        
        subs.push_back(vector(1, nums[0]));
        for (int i = 1; i < length; i ++)
        {
            int minLen = INT_MAX;
            int index = -1;
            for (int j = 0; j < subs.size(); j ++)
            {
                if (nums[i] == subs[j].back() + 1)
                {
                    if (subs[j].size() < minLen)
                    {
                        index = j;
                        minLen = subs[j].size();
                    }
                }
            }
            if (index == -1)
            {
                subs.push_back(vector(1, nums[i]));
                continue;
            }
            else
            {
                subs[index].push_back(nums[i]);
            }
        }
        
        for (int i = 0; i < subs.size(); i ++)
        {
            if (subs[i].size() < 3)
                return false;
        }
        return true;
    }
};

Accept解法

class Solution {
public:
    bool isPossible(vector& nums) {
        map freq;
        map q;
        
        int length = nums.size();
        for (int i = 0; i < length; i ++)
        {
            if (freq.find(nums[i]) == freq.end())
                freq[nums[i]] = 1;
            else
                freq[nums[i]] ++;
        }
        
        for (int i = 0; i < nums.size(); i ++)
        {
            int current = nums[i];
            if (freq[current] == 0)
            {
                continue;
            }
            if (q.find(current) == q.end() || q[current] == 0)
            {
                if (freq.find(current + 1) == freq.end() ||
                    freq.find(current + 2) == freq.end() ||
                    freq[current + 1] == 0 ||
                    freq[current + 2] == 0)
                {
                    return false;
                }
                if (q.find(current + 3) == q.end())
                    q.insert(make_pair(current + 3, 1));
                else
                    q[current + 3] ++;
                freq[current] --;
                freq[current + 1] --;
                freq[current + 2] --;
            }
            else
            {
                freq[current] --;
                q[current] --;
                if (q.find(current + 1) == q.end())
                    q[current + 1] = 1;
                else
                    q[current + 1] ++;
            }
        }
        
        return true;
    }
};

运行效率

Runtime: 188 ms, faster than 20.00% of C++ online submissions for Split Array into Consecutive Subsequences.

Memory Usage: 18.1 MB, less than 93.55% of C++ online submissions forSplit Array into Consecutive Subsequences.