[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)

kotlin, O(2^n)