Combination Sum 系列题解

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];
}
};

进阶(Combination Sum II)

进阶版本在第一版的基础上对增加了更多的条件:给出的candidates数组的元素是可以重复的,所以最大问题就是去重,下面给出递归DFS的做法:

class Solution {
private:
int size;
void dfs(vector<vector<int>>& res, vector<int>& path,
vector<int>& candidates, int begin, int target) {
if (target == 0) {
res.push_back(path);
} else if (begin < size) {
for (int i = begin; i < size; i++) {
if (candidates[i] <= target) {
if (i > 0 && i > begin &&
candidates[i] == candidates[i - 1])
continue; //去重的关键过滤步骤:跳过所有相同的元素
path.push_back(candidates[i]);
dfs(res, path, candidates,
i + 1, target - candidates[i]);
path.pop_back();
} else {
return;
}
}
}
}
public:
vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
sort(candidates.begin(), candidates.end());
this->size = candidates.size();
vector<vector<int>> res;
vector<int> path;
dfs(res, path, candidates, 0, target);
return res;
}
};

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