[LeetCode] 39. Combination Sum

轉自LeetCode

Given a set of candidate numbers (C) 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.

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is: 

[
  [7],
  [2, 2, 3]
]

<Solution>

要求排列組合，並且有個最終滿足條件 => DFS

這題要注意一個地方是，數字可以重複用，這點處理好就可以了

code 如下
c++

	class Solution {
	public:
	vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
	vector<vector<int>> ans;
	vector<int> out;
	//> use DFS
	combineByDFS(0, target, candidates, out, ans);
	return ans;
	}
	void combineByDFS(const int &start, const int &target, vector<int> &candidates, vector<int> &out, vector<vector<int>> &ans) {
	if(target == 0) {
	//> find answer
	ans.push_back(out);
	}
	else {
	for(int i = start; i < candidates.size(); i++) {
	if((target - candidates[i]) >= 0) {
	out.push_back(candidates[i]);
	//> start = i, because the same value can be used multi-times
	combineByDFS(i, target-candidates[i], candidates, out, ans);
	out.pop_back();
	}
	}
	}
	}
	};

view raw combinationSum.cpp hosted with ❤ by GitHub

kotlin

	class Solution {
	fun combinationSum(candidates: IntArray, target: Int): List<List<Int>> {
	val ans = mutableListOf<List<Int>>()
	dfs(0, candidates, target, emptyList(), ans)
	return ans
	}
	fun dfs(start: Int, candidates: IntArray, target: Int, cmb: List<Int>, ans: MutableList<List<Int>>) {
	if (target == 0) {
	ans.add(cmb)
	return
	}
	for(i in start..candidates.lastIndex) {
	if(target - candidates[i] >= 0) {
	dfs(i, candidates, target - candidates[i], cmb+listOf(candidates[i]), ans)
	}
	}
	}
	}

view raw combination_sum.kt hosted with ❤ by GitHub