491. Increasing Subsequences

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.

题目含义：给定几个整数，求他们能构造的所有递增子序列

    void increSubSeq(int[] nums, Set<List<Integer>> result, List<Integer> tmp, int nextIndex) {
        if (tmp.size() >= 2) result.add(new LinkedList<>(tmp));
        for (int i = nextIndex; i < nums.length; i++) {
            if (tmp.size() == 0 || nums[i] >= tmp.get(tmp.size()-1)) {
                tmp.add(nums[i]);
                increSubSeq(nums, result, tmp, i + 1);
                tmp.remove(tmp.size()-1);//将nums[nextIndex]弹出，继续下一次dfs。
            }
        }
    }
    
    public List<List<Integer>> findSubsequences(int[] nums) {
//        这道题可以采用深度优先搜索的方法解决。假设使用tmp来存储当前考虑的子序列，nextIndex表示tmp中最后一个元素的下标的后一个，
//        则如果nums[nextIndex]的元素不小于tmp中的最后一个元素，就可以将其增加到tmp中，然后继续处理nextIndex+1处的元素，直到最后一个元素或者元素比最后一个小。
//        接着返回原始的tmp，将nums[nextIndex]弹出，继续下一次dfs。
        Set<List<Integer>> result = new HashSet<>();
        List<Integer> tmp = new LinkedList<>();
        increSubSeq(nums, result, tmp, 0);
        return new ArrayList<>(result);
    }

方法二: 时间复杂度不符合要求

    public List<List<Integer>> findSubsequences(int[] nums) {
        if (nums.length == 0) return new ArrayList<>();
        Set<List<Integer>> result = new HashSet<>();
        List<List<Integer>> list = new LinkedList<>();
        list.add(new ArrayList<>());
        for (int i = 0; i < nums.length;i++)
        {
            int size = list.size();
            for (int j=0;j<size;j++)
            {
                List<Integer> tempList = new ArrayList<>(list.get(j));
                if (tempList.size()>0 && tempList.get(tempList.size()-1) > nums[i]) continue;
                tempList.add(nums[i]);
                list.add(tempList);
                if (tempList.size()>1) result.add(tempList);
            }
        }
        return new ArrayList<>(result);
    }