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Subject: 491. Increasing Subsequence
Given an integer array, your task is to find all increasing subsequences of the array, and the length of the increasing subsequence is at least 2.
Example:
输入: [4, 6, 7, 7]
输出: [[4, 6], [4, 7], [4, 6, 7], [4, 6, 7, 7], [6, 7], [6, 7, 7], [7,7], [4,7,7]]
Description:
- The length of the given array will not exceed 15.
- The range of integers in the array is [-100,100].
- A given array may contain repeated numbers, and equal numbers should be considered as an increase.
Source: LeetCode
Link: https://leetcode-cn.com/problems/increasing-subsequences
Copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.
Basic idea: dfs
Here to explain: the title requires an increasing subsequence, the only difference between a subsequence and a substring is: the subsequence can be discontinuous. It is obvious here: the subsequence should maintain the relative order of each number in the original array.
Deduplication with the help of set:
class Solution {
public:
set<vector<int>> res;
vector<vector<int>> findSubsequences(vector<int>& nums) {
vector<vector<int>> vres;
if(nums.size() < 2)
return vres;
//子序列要保持相对顺序
vector<int> cur;
dfs(nums, 0, cur);
for(auto r : res){
vres.push_back(r);
}
return vres;
}
void dfs(vector<int>& nums, int pos, vector<int> cur){
if(cur.size() > 1)
res.insert(cur);
for(int i = pos; i < nums.size(); ++i){
if(cur.size() == 0 || nums[i] >= cur.back()){
cur.push_back(nums[i]);
dfs(nums, i + 1, cur);
cur.pop_back();
}
}
}
};
Pruning and de-duplication in the process of dfs: It is
realized by saving the popped elements when backtracking, to ensure that the popped elements will not enter the current array again.
class Solution {
public:
vector<vector<int>> res;
vector<vector<int>> findSubsequences(vector<int>& nums) {
if(nums.size() < 2)
return res;
vector<int> cur;
dfs(nums, 0, cur);
return res;
}
void dfs(vector<int>& nums, int pos, vector<int> cur){
if(cur.size() > 1)
res.push_back(cur);
unordered_set<int> help;
for(int i = pos; i < nums.size(); ++i){
if((cur.size() == 0 || nums[i] >= cur.back()) && help.find(nums[i]) == help.end()){
cur.push_back(nums[i]);
dfs(nums, i + 1, cur);
cur.pop_back();
//去掉重复的元素
help.insert(nums[i]);
}
}
}
};