[LeetCode] 491. Increasing Subsequences

最新推荐文章于 2025-01-18 01:32:59 发布

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最新推荐文章于 2025-01-18 01:32:59 发布

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原文链接：https://segmentfault.com/a/1190000017138824

本文介绍了一种寻找所有可能的递增子序列的算法，输入为整数数组，输出长度至少为2的所有不同递增子序列。示例中输入[4,6,7,7]，输出包括[4,6]、[4,7]等在内的8个子序列。算法通过深度优先搜索实现，避免重复并考虑相等整数作为递增序列的特殊情况。

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Problem

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.

Solution

class Solution {
    public List<List<Integer>> findSubsequences(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if (nums == null || nums.length == 0) return res;
        dfs(nums, 0, new ArrayList<Integer>(), res);
        return res;
    }
    private void dfs(int[] nums, int start, List<Integer> temp, List<List<Integer>> res) {
        if (temp.size() > 1) {
            res.add(new ArrayList<>(temp));
        }
        Set<Integer> used = new HashSet<>();
        for (int i = start; i < nums.length; i++) {
            if (used.contains(nums[i])) continue;
            if (temp.size() == 0 || nums[i] >= temp.get(temp.size()-1)) {
                temp.add(nums[i]);
                used.add(nums[i]);
                dfs(nums, i+1, temp, res); //next dfs doesn't have used, yeah
                temp.remove(temp.size()-1);
            }
        }
    }
}