[LeetCode] (medium) 491. Increasing Subsequences-优快云博客

本文介绍了一种使用深度优先搜索（DFS）算法解决LeetCode上的一道题目：寻找所有可能的递增子序列，子序列长度至少为2。通过一个示例展示了输入[4,6,7,7]时的输出结果，并详细解释了如何通过剪枝策略避免重复子序列，确保算法效率。

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https://leetcode.com/problems/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:

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.

dfs，针对每一个层级内部记录并排除重复实现剪枝，前面的子串对后面的子串有完全的覆盖效果（在同一个层级中）

class Solution {
public:
    vector<vector<int>> result;
    vector<int> inter;
    
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        dfs(0, nums);
        
        return result;
    }
    
    void dfs(int pos, vector<int>& nums){
        if(inter.size() >= 2) result.push_back(inter);
        if(pos >= nums.size()) return;
        
        unordered_map<int, int> used;
        
        for(int i = pos; i < nums.size(); ++i){
            if((inter.empty() || nums[i] >= inter.back()) && used.find(nums[i]) == used.end()){
                inter.push_back(nums[i]);
                dfs(i+1, nums);
                inter.pop_back();
                used[nums[i]] = 1;
            }
        }
    }
};