40 Combination Sum II – Medium

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. Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak). 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]

Thoughts:

This is similar to the I version of the problem. Only difference is that the element in the collection can only be used once.

The index passed in is different and need to check num[i] == num[i-1] .

if (i == j || candidates[i] != candidates[i-1]) then continue to go deeper down.

Solutions:

``````public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(candidates == null || candidates.length == 0) return result;
ArrayList<Integer> current = new ArrayList<Integer>();
Arrays.sort(candidates);
combinationSum(candidates, target, 0, current, result);
return result;
}
public void combinationSum(int[] candidates, int target, int j, ArrayList<Integer> curr, List<List<Integer>> result){
if(target == 0){
ArrayList<Integer> temp = new ArrayList<Integer>(curr);
return;
}
for(int i=j; i<candidates.length; i++){
if (target < candidates[i]){
return;
}
if (i == j || candidates[i] != candidates[i-1]){