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1. 题意

给定一个数组，和一个目标值；求得所有数组中所有和为目标值的元素序列。

组合总数

2. 题解

回溯列举每一个可能的序列，注意去重。

2.1 我的解法

class Solution {
public:void gen(vector<vector<int>> &ans,const vector<int> &candidates, vector<int> &seq, int target){if (target == 0) {ans.push_back(seq);return ;}if ( target < 0)return ;int sz = candidates.size();for ( int i = 0; i < sz; ++i) {if ( !seq.empty() &&candidates[i] < seq[seq.size() - 1])continue;seq.push_back(candidates[i]);gen(ans, candidates, seq, target - candidates[i]);seq.pop_back();}}vector<vector<int>> combinationSum(vector<int>& candidates, int target) {vector<vector<int>> ans;vector<int> one;sort(candidates.begin(), candidates.end());gen(ans, candidates, one, target);return ans;}
};

2.2 另一种可能

class Solution {
public:void gen(vector<vector<int>> &ans,const vector<int> &candidates, vector<int> &seq, int idx, int target){if (target == 0) {ans.push_back(seq);return ;}if ( target < 0)return ;int sz = candidates.size();for ( int i = idx; i < sz; ++i) {seq.push_back(candidates[i]);gen(ans, candidates, seq, i, target - candidates[i]);seq.pop_back();}}vector<vector<int>> combinationSum(vector<int>& candidates, int target) {vector<vector<int>> ans;vector<int> one;sort(candidates.begin(), candidates.end());gen(ans, candidates, one, 0, target);return ans;}
};