Increasing Subsequences

寻找递增子序列

最新推荐文章于 2020-04-05 12:00:37 发布

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最新推荐文章于 2020-04-05 12:00:37 发布

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分类专栏： leetcode ACM 文章标签： Increasing Subsequen

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本文介绍了一种使用深度优先搜索（DFS）算法来查找给定整数数组中所有不同的递增子序列的方法。通过实例展示了如何避免重复子序列并确保递增子序列长度至少为2。

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.

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解析：

求所有的递增序列， dfs(vector<vector<int>>&ans,vector<int>&path,vector<int>nums,int begin) 表示从begin开始的所有递增序列，为了防止存在重复的子序列，定义一个vis表示前面相同的元素是否已经添加进去了。

代码：

class Solution {
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        vector<vector<int>>ans;
        vector<int>path;
        dfs(ans,path,nums,0);
        return ans;
    }
    
    void dfs(vector<vector<int>>&ans,vector<int>&path,vector<int>nums,int begin)
    {
        if (path.size()>=2)
        {
            ans.push_back(path);
        }
        
        
        vector<bool>vis(220,false);
        for (int i=begin; i<nums.size(); i++)
        {
            
            if (path.empty()||nums[i]>=(path[path.size()-1]))
            {
                if (vis[nums[i]+110])
                continue;
                vis[nums[i]+110]=true;
                
                path.push_back(nums[i]);
                dfs(ans,path,nums,i+1);
                path.pop_back();
            }
        }
        return ;
    }
    
};