491. Increasing Subsequences

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BigFatSheep

最新推荐文章于 2024-01-20 15:08:16 发布

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本文介绍了一种使用递归辅助函数查找整数数组中所有不同递增子序列的方法。通过避免重复情况并确保子序列长度至少为2，该算法有效地解决了问题。适用于长度不超过15的数组，范围在[-100,100]之间。

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Description

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

Problem URL

Solution

给一个整数数组，找到这个数组中所有的递增子序列。

Using a helper function to recursively find all the subsequence. If the subquence’s length is longer than 1, add it into result list. Then use a for loop to find subsequences. A hash set is used to avoid duplicate circumstance.

Code

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