[LeetCode] 40. Combination Sum II

轉自LeetCode

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• The solution set must not contain duplicate combinations.

For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is: 

[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

<Solution>

這題是 Combination Sum 的衍生，一樣要求不重複的排列組合的和

差別在於，每個數字只能用一次

可以基於之前的解法，修改一下來避免重複

• 為了要避開重複，先 sort

• 因為每個數字只能用一次，所以每次的 recursive， start = i+1

• 在 DFS 的 for loop 裡面，增加以下片段來避免重複

if(i > start && candidates[i] == candidates[i-1]) {
    continue;
}

code 如下
c++, O(2^n)

	class Solution {
	public:
	vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
	vector<vector<int>> ans;
	vector<int> out;
	//> to avoid duplicate, sort first
	sort(candidates.begin(), candidates.end());
	//> use DFS to solve
	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) {
	ans.push_back(out);
	}
	else {
	for(int i = start; i < candidates.size(); i++) {
	//> avoid duplicates
	if(i > start && candidates[i] == candidates[i-1]) {
	continue;
	}
	else if((target-candidates[i]) >= 0) {
	out.push_back(candidates[i]);
	//> start = i+1, use each number once
	combineByDFS(i+1, target-candidates[i], candidates, out, ans);
	out.pop_back();
	}
	}
	}
	}
	};

view raw combinationSum_2.cpp hosted with ❤ by GitHub

kotlin, O(2^n)

	class Solution {
	private val ans = arrayListOf<List<Int>>()
	fun combinationSum2(candidates: IntArray, target: Int): List<List<Int>> {
	dfs(0, candidates.sorted(), target, emptyList())
	return ans
	}
	fun dfs(start: Int, candidates: List<Int>, target: Int, cmb: List<Int>) {
	if (target == 0) {
	ans.add(cmb)
	return
	}
	for(i in start until candidates.size) {
	if (i > start && candidates[i] == candidates[i-1]) {
	continue
	}
	if (target - candidates[i] >= 0) {
	dfs(i+1, candidates, target - candidates[i], cmb + listOf(candidates[i]))
	}
	}
	}
	}

view raw combination_sum_II.kt hosted with ❤ by GitHub