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  • 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.

    Approach #1: DFS. [Java]

    class Solution {
        public List<List<Integer>> findSubsequences(int[] nums) {
            List<List<Integer>> res = new LinkedList<>();
            helper(new LinkedList<Integer>(), 0, nums, res);
            return res;
        }
        
        
        public void helper(LinkedList<Integer> list, int index, int[] nums, List<List<Integer>> res) {
            if (list.size() > 1) res.add(new LinkedList<Integer>(list));
            Set<Integer> used = new HashSet<>();
            for (int i = index; i < nums.length; ++i) {
                if (used.contains(nums[i])) continue;
                if (list.size() == 0 || nums[i] >= list.peekLast()) {
                    used.add(nums[i]);
                    list.add(nums[i]);
                    helper(list, i+1, nums, res);
                    list.remove(list.size() - 1);
                }
            }
        }
    }
    

      

    Analysis:

    Using the list.size() as the foundation to process dfs.

    Reference:

    https://leetcode.com/problems/increasing-subsequences/discuss/97147/Java-solution-beats-100

    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/10957852.html
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