[LeetCode] 491. Increasing Subsequences 遞增子序列


 

Given an integer array, your task is to find all the different possible increasing subsequences of the given array, and the length of an increasing subsequence should be at least 2 .

Example:

Input: [4, 6, 7, 7]
Output: [[4, 6], [4, 7], [4, 6, 7], [4, 6, 7, 7], [6, 7], [6, 7, 7], [7,7], [4,7,7]]

 

Note:

  1. The length of the given array will not exceed 15.
  2. The range of integer in the given array is [-100,100].
  3. The given array may contain duplicates, and two equal integers should also be considered as a special case of increasing sequence.

 

這道題讓我們找出所有的遞增子序列,應該不難想到,這題肯定是要先找出所有的子序列,從中找出遞增的。找出所有的子序列的題之前也接觸過 Subsets 和 Subsets II,那兩題不同之處在於數組中有沒有重復項。而這道題明顯是有重復項的,所以需要用到 Subsets II 中的解法。首先來看一種迭代的解法,對於重復項的處理,最偷懶的方法是使用 TreeSet,利用其自動去處重復項的機制,然后最后返回時再轉回 vector 即可。由於是找遞增序列,所以需要對遞歸函數做一些修改,首先題目中說明了遞增序列數字至少兩個,所以只有子序列個數大於等於2時,才加入結果。然后就是要遞增,如果之前的數字大於當前的數字,那么跳過這種情況,繼續循環,參見代碼如下:

 

解法一:

class Solution {
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        set<vector<int>> res;
        vector<int> out;
        helper(nums, 0, out, res);
        return vector<vector<int>>(res.begin(), res.end());
    }
    void helper(vector<int>& nums, int start, vector<int>& out, set<vector<int>>& res) {
        if (out.size() >= 2) res.insert(out);
        for (int i = start; i < nums.size(); ++i) {
            if (!out.empty() && out.back() > nums[i]) continue;
            out.push_back(nums[i]);
            helper(nums, i + 1, out, res);
            out.pop_back();
        }
    }
};

 

我們也可以在遞歸中進行去重復處理,方法是用一個 HashSet 保存中間過程的數字,如果當前的數字在之前出現過了,就直接跳過這種情況即可,參見代碼如下:

 

解法二:

class Solution {
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> out;
        helper(nums, 0, out, res);
        return res;
    }
    void helper(vector<int>& nums, int start, vector<int>& out, vector<vector<int>>& res) {
        if (out.size() >= 2) res.push_back(out);
        unordered_set<int> st;
        for (int i = start; i < nums.size(); ++i) {
            if ((!out.empty() && out.back() > nums[i]) || st.count(nums[i])) continue;
            out.push_back(nums[i]);
            st.insert(nums[i]);
            helper(nums, i + 1, out, res);
            out.pop_back();
        }
    }
};

 

下面我們來看迭代的解法,還是老套路,先看偷懶的方法,用 TreeSet 來去處重復。對於遞歸的處理方法跟之前相同,參見代碼如下:

 

解法三:

class Solution {
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        set<vector<int>> res;
        vector<vector<int>> cur(1);
        for (int i = 0; i < nums.size(); ++i) {
            int n = cur.size();
            for (int j = 0; j < n; ++j) {
                if (!cur[j].empty() && cur[j].back() > nums[i]) continue;
                cur.push_back(cur[j]);
                cur.back().push_back(nums[i]);
                if (cur.back().size() >= 2) res.insert(cur.back());
            }
        }
        return vector<vector<int>>(res.begin(), res.end());
    }
};

 

我們來看不用 TreeSet 的方法,使用一個 HashMap 來建立每個數字對應的遍歷起始位置,默認都是0,然后在遍歷的時候先取出原有值當作遍歷起始點,然后更新為當前位置,如果某個數字之前出現過,那么取出的原有值就不是0,而是之前那個數的出現位置,這樣就不會產生重復了,如果不太好理解的話就帶個簡單的實例去試試吧,參見代碼如下:

 

解法四:

class Solution {
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<vector<int>> res, cur(1);
        unordered_map<int, int> m;
        for (int i = 0; i < nums.size(); ++i) {
            int n = cur.size(), start = m[nums[i]];
            m[nums[i]] = n;
            for (int j = start; j < n; ++j) {
                if (!cur[j].empty() && cur[j].back() > nums[i]) continue;
                cur.push_back(cur[j]);
                cur.back().push_back(nums[i]);
                if (cur.back().size() >= 2) res.push_back(cur.back());
            }
        }
        return res;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/491

 

類似題目:

Subsets

Subsets II

Maximum Length of Pair Chain

 

參考資料:

https://leetcode.com/problems/increasing-subsequences/

https://leetcode.com/problems/increasing-subsequences/discuss/97124/c-dfs-solution-using-unordered_set

https://leetcode.com/problems/increasing-subsequences/discuss/97134/evolve-from-intuitive-solution-to-optimal

 

LeetCode All in One 題目講解匯總(持續更新中...)


免責聲明!

本站轉載的文章為個人學習借鑒使用,本站對版權不負任何法律責任。如果侵犯了您的隱私權益,請聯系本站郵箱yoyou2525@163.com刪除。



 
粵ICP備18138465號   © 2018-2025 CODEPRJ.COM