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feat: add solutions to lc/lcof2 problem
lc No.0039 & lcof2 No.081. Combination Sum
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‎lcof2/剑指 Offer II 081. 允许重复选择元素的组合/README.md

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<p><meta charset="UTF-8" />注意：本题与主站 39&nbsp;题相同：&nbsp;<a href="https://leetcode-cn.com/problems/combination-sum/">https://leetcode-cn.com/problems/combination-sum/</a></p>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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DFS。
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为了避免重复方案，需要定义一个搜索起点。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
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ans = []
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n = len(candidates)
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def dfs(s, u, t):
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if s == target:
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ans.append(t.copy())
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return
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if s > target:
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return
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for i in range(u, n):
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c = candidates[i]
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t.append(c)
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dfs(s + c, i, t)
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t.pop()
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dfs(0, 0, [])
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private List<List<Integer>> ans;
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private int target;
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private int[] candidates;
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public List<List<Integer>> combinationSum(int[] candidates, int target) {
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ans = new ArrayList<>();
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this.target = target;
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this.candidates = candidates;
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dfs(0, 0, new ArrayList<>());
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return ans;
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}
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private void dfs(int s, int u, List<Integer> t) {
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if (s == target) {
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ans.add(new ArrayList<>(t));
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return;
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}
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if (s > target) {
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return;
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}
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for (int i = u; i < candidates.length; ++i) {
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int c = candidates[i];
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t.add(c);
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dfs(s + c, i, t);
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t.remove(t.size() - 1);
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}
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<vector<int>> ans;
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vector<int> candidates;
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int target;
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vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
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this->candidates = candidates;
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this->target = target;
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vector<int> t;
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dfs(0, 0, t);
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return ans;
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}
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void dfs(int s, int u, vector<int>& t) {
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if (s == target)
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{
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ans.push_back(t);
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return;
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}
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if (s > target) return;
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for (int i = u; i < candidates.size(); ++i)
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{
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int c = candidates[i];
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t.push_back(c);
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dfs(s + c, i, t);
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t.pop_back();
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}
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}
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};
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```
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### **Go**
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```go
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func combinationSum(candidates []int, target int) [][]int {
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var ans [][]int
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var dfs func(s, u int, t []int)
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dfs = func(s, u int, t []int) {
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if s == target {
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ans = append(ans, append([]int(nil), t...))
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return
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}
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if s > target {
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return
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}
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for i := u; i < len(candidates); i++ {
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c := candidates[i]
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t = append(t, c)
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dfs(s+c, i, t)
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t = t[:len(t)-1]
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}
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}
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var t []int
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dfs(0, 0, t)
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return ans
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}
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```
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### **...**

‎lcof2/剑指 Offer II 081. 允许重复选择元素的组合/Solution.cpp

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class Solution {
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public:
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vector<vector<int>> ans;
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vector<int> candidates;
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int target;
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vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
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this->candidates = candidates;
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this->target = target;
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vector<int> t;
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dfs(0, 0, t);
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return ans;
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}
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void dfs(int s, int u, vector<int>& t) {
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if (s == target)
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{
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ans.push_back(t);
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return;
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}
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if (s > target) return;
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for (int i = u; i < candidates.size(); ++i)
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{
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int c = candidates[i];
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t.push_back(c);
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dfs(s + c, i, t);
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t.pop_back();
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}
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}
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};

‎lcof2/剑指 Offer II 081. 允许重复选择元素的组合/Solution.cs

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using System;
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using System.Collections.Generic;
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using System.Linq;
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public class Solution
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{
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public IList<IList<int>> CombinationSum(int[] candidates, int target)
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{
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Array.Sort(candidates);
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candidates = candidates.Distinct().ToArray();
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var paths = new List<int>[target + 1];
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paths[0] = new List<int>();
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foreach (var c in candidates)
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{
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for (var j = c; j <= target; ++j)
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{
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if (paths[j - c] != null)
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{
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if (paths[j] == null)
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{
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paths[j] = new List<int>();
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}
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paths[j].Add(c);
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}
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}
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}
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var results = new List<IList<int>>();
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if (paths[target] != null) GenerateResults(results, new Stack<int>(), paths, target, paths[target].Count - 1);
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return results;
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}
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private void GenerateResults(IList<IList<int>> results, Stack<int> result, List<int>[] paths, int remaining,
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int maxIndex)
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{
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if (remaining == 0)
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{
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results.Add(new List<int>(result));
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return;
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}
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for (var i = maxIndex; i >= 0; --i)
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{
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var value = paths[remaining][i];
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result.Push(value);
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var nextMaxIndex = paths[remaining - value].BinarySearch(value);
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if (nextMaxIndex < 0)
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{
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nextMaxIndex = ~nextMaxIndex - 1;
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}
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GenerateResults(results, result, paths, remaining - value, nextMaxIndex);
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result.Pop();
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}
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}
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}

‎lcof2/剑指 Offer II 081. 允许重复选择元素的组合/Solution.go

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func combinationSum(candidates []int, target int) [][]int {
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var ans [][]int
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var dfs func(s, u int, t []int)
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dfs = func(s, u int, t []int) {
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if s == target {
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ans = append(ans, append([]int(nil), t...))
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return
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}
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if s > target {
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return
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}
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for i := u; i < len(candidates); i++ {
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c := candidates[i]
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t = append(t, c)
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dfs(s+c, i, t)
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t = t[:len(t)-1]
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}
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}
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var t []int
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dfs(0, 0, t)
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return ans
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}

‎lcof2/剑指 Offer II 081. 允许重复选择元素的组合/Solution.java

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class Solution {
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private List<List<Integer>> ans;
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private int target;
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private int[] candidates;
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public List<List<Integer>> combinationSum(int[] candidates, int target) {
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ans = new ArrayList<>();
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this.target = target;
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this.candidates = candidates;
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dfs(0, 0, new ArrayList<>());
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return ans;
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}
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private void dfs(int s, int u, List<Integer> t) {
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if (s == target) {
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ans.add(new ArrayList<>(t));
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return;
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}
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if (s > target) {
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return;
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}
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for (int i = u; i < candidates.length; ++i) {
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int c = candidates[i];
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t.add(c);
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dfs(s + c, i, t);
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t.remove(t.size() - 1);
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}
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}
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}

‎lcof2/剑指 Offer II 081. 允许重复选择元素的组合/Solution.py

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class Solution:
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def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
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ans = []
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n = len(candidates)
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def dfs(s, u, t):
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if s == target:
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ans.append(t.copy())
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return
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if s > target:
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return
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for i in range(u, n):
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c = candidates[i]
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t.append(c)
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dfs(s + c, i, t)
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t.pop()
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dfs(0, 0, [])
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return ans

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