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，那麼我們應該不難想到，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我們之前也接觸過

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();
        }
    }
};

我們也可以在遞迴中進行去重複處理，方法是用一個set儲存中間過程的數字，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，參見程式碼如下：

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set來去處重複。對於遞迴的處理方法跟之前相同，參見程式碼如下：

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set的方法，使用一個雜湊表來建立每個數字對應的遍歷起始位置，預設都是0，然後在遍歷的時候先取出原有值當作遍歷起始點，然後更新為當前位置，如果某個數字之前出現過，那麼取出的原有值就不是0，而是之前那個數的出現位置，這樣就就不會產生重複了，如果不太好理解的話就帶個簡單的例項去試試吧，參見程式碼如下：

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();
            int 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;
    }
};

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