Leetcode: Combination Sum II

Problem

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:

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

Leetcode Link

https://leetcode.com/problems/combination-sum-ii/#/description

Solution

This problem can be solved using DFS:

  • Get the result starting with the first number
  • Get the result starting with the second number
  • Get the result starting with the last number

But we need to sort the array first in order to remove duplicate records.

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public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
if(candidates == null || candidates.length == 0) {
return new ArrayList<>();
}
Arrays.sort(candidates);
return combinationSum2(candidates, target, new ArrayList<>(), 0);
}
private List<List<Integer>> combinationSum2(int[] candidates, int target, List<Integer> prefix, int startPos) {
List<List<Integer>> result = new ArrayList<>();
if(target == 0) {
result.add(new ArrayList<>(prefix));
} else if(target > 0){
for (int i = startPos; i < candidates.length; i++) {
if(i > startPos && candidates[i] == candidates[i - 1]) {
continue;
}
prefix.add(candidates[i]);
List<List<Integer>> subResult = combinationSum2(candidates, target - candidates[i], prefix, i + 1);
prefix.remove(prefix.size() - 1);
result.addAll(subResult);
}
}
return result;
}
}
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