Combination Sum 题解
题目来源：https://leetcode.com/problems/combination-sum/description/
Description
Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
The same repeated number may be chosen from C unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- The solution set must not contain duplicate combinations.
Example
For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:
[
[7],
[2, 2, 3]
]Solution
class Solution {
private:
void backTrack(vector<int>& path, vector<vector<int> >& res,
vector<int>& candidates, int begin, int target) {
if (target == 0) {
res.push_back(path);
} else {
int size = candidates.size();
if (begin >= size)
return;
for (int i = begin; i < size; i++) {
if (candidates[i] <= target) {
path.push_back(candidates[i]);
backTrack(path, res, candidates, i, target - candidates[i]);
path.pop_back();
}
}
}
}
public:
vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
vector<vector<int> > res;
vector<int> path;
backTrack(path, res, candidates, 0, target);
return res;
}
};解题描述
这道题类似与经典的零钱兑换问题，在给定的数组candidates，找出所有和为target的数字组合，选择的数字可以重复但解法不能重复。上面使用的是递归回溯的办法，类似DFS，不难理解，下面再多给出使用迭代DP的解法：
class Solution {
public:
vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
vector<vector<vector<int> > > dp(target + 1, vector<vector<int> >());
dp[0].push_back(vector<int>());
for (auto candidate : candidates) {
for (int j = candidate; j <= target; j++) {
if (!dp[j - candidate].empty()) {
auto paths = dp[j - candidate];
for (auto& path : paths)
path.push_back(candidate);
dp[j].insert(dp[j].end(), paths.begin(), paths.end());
}
}
}
return dp[target];
}
};