[LeetCode] Combination Sum 2

Question

Analysis

This question is similar to the question Path Sum 2 and Two Sum. I choose recursion to solve it, just like Path Sum 2.
The main idea is easy, but an important question is how to eliminate duplicate.It’s the difficultest part of this question. I use the idea that it cannot appear the same value at the same index(the relative position of the array ). I just need to add an condition sentence before entering recursive sentence.

The analysis of the time complexity

Exactly, I use the idea of dfs, but I am not sure about it’s time complexity. I think it’s O(n2).

Code

#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
    vector<vector<int> > combinationSum2(vector<int>& candidates, int target) {
        sort(candidates.begin(), candidates.end());
        can = candidates;
        size = candidates.size();
        vector<int> temp;
        for (int i = 0; i < size; i++) {
            if (i > 0 && can[i -1] == can[i]) continue;
            findOneAnswer(temp, i, target);
        }
        return ans;
    }
    void findOneAnswer(vector<int> & v, int index, int target) {
        if (can[index] > target) return;
        v.push_back(can[index]);
        if (target - can[index] == 0) {
            ans.push_back(v);
            v.pop_back();
            return;
        }
        for (int i = index + 1; i < size; i++) {
            if (i > index + 1 && can[i -1] == can[i]) continue;
            findOneAnswer(v, i, target - can[index]);
        }
        v.pop_back();
    }
    vector<vector<int> > ans;
    vector<int> can;
    int size;
};