[LeetCode] Combination Sum

Combination Sum

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.
• 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 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3] 

Hashtags:  Array Backtracking

Solution: We first sort the array (for the 2nd constraint in the problem description).
For the 1st element a, we use 0, 1, ...., t number of a's, where t = target / a. Then we solve the problem recursively using the remaining n - 1 elements and updated target value. 

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public class Solution { private List<List<Integer>> combinationSum( int[] candidates, int i, int target ) { ArrayList<List<Integer>> list = new ArrayList<List<Integer>>(); if (i >= candidates.length) { return list; } int val = candidates[i]; ArrayList<Integer> tmp = new ArrayList<Integer>(); for (int j = 0; j <= target / val; j++) { int new_target = target - j * val; if (new_target == 0) { list.add(tmp); continue; } List<List<Integer>> ll = combinationSum( candidates, i + 1, new_target ); for (List<Integer> li : ll) { ArrayList<Integer> l = new ArrayList<Integer>(tmp); l.addAll(li); list.add(l); } tmp.add(val); } return list; } public List<List<Integer>> combinationSum( int[] candidates, int target ) { Arrays.sort(candidates); return combinationSum(candidates, 0, target); } }