文章作者:Tyan
博客:noahsnail.com | 优快云 | 简书
1. Description

2. Solution
- Version 1
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
set<vector<int>> s;
sort(candidates.begin(), candidates.end());
vector<int> combination;
combinationSum2(s, candidates, combination, target, 0, 0);
return vector<vector<int>>(s.begin(), s.end());
}
private:
void combinationSum2(set<vector<int>>& result, vector<int>& candidates, vector<int>& combination, int& target, int sum, int begin) {
if(sum > target) {
return;
}
if(sum == target) {
result.insert(combination);
return;
}
for(int i = begin; i < candidates.size(); i++) {
combination.push_back(candidates[i]);
combinationSum2(result, candidates, combination, target, sum + candidates[i], i + 1);
combination.pop_back();
}
}
};
- Version 2
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
set<vector<int>> s;
sort(candidates.begin(), candidates.end());
vector<int> combination;
combinationSum2(s, candidates, combination, target, 0, 0);
return vector<vector<int>>(s.begin(), s.end());
}
private:
void combinationSum2(set<vector<int>>& result, vector<int>& candidates, vector<int>& combination, int& target, int sum, int begin) {
if(sum > target) {
return;
}
if(sum == target) {
result.insert(combination);
return;
}
for(int i = begin; i < candidates.size(); i++) {
if(sum + candidates[i] > target) {
break;
}
combination.push_back(candidates[i]);
combinationSum2(result, candidates, combination, target, sum + candidates[i], i + 1);
combination.pop_back();
}
}
};
- Version 3
class Solution {
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
vector<vector<int>> result;
sort(candidates.begin(), candidates.end());
vector<int> combination;
combinationSum2(result, candidates, combination, target, 0, 0);
return result;
}
private:
void combinationSum2(vector<vector<int>>& result, vector<int>& candidates, vector<int>& combination, int& target, int sum, int begin) {
if(sum > target) {
return;
}
if(sum == target) {
result.push_back(combination);
return;
}
for(int i = begin; i < candidates.size(); i++) {
if(sum + candidates[i] > target) {
break;
}
if(i > begin && candidates[i] == candidates[i - 1]) {
continue;
}
combination.push_back(candidates[i]);
combinationSum2(result, candidates, combination, target, sum + candidates[i], i + 1);
combination.pop_back();
}
}
};

本文提供了三种使用C++解决LeetCode上组合总和II问题的不同版本代码实现,详细展示了如何通过递归和回溯算法寻找候选数集中所有可能的组合,使得组合之和等于目标值,同时避免重复组合。
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