LeetCode - Combination Sum
Problem description
Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.
The same repeated number may be chosen from candidates unlimited number of times.
Note:
- All numbers (including target) will be positive integers.
- The solution set must not contain duplicate combinations.
Example 1:
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Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
[7],
[2,2,3]
]
Example 2:
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Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
Analysis
This is a standard backtrack. Here’s the code:
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public class CombinationSum {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
Arrays.sort(candidates);
List<List<Integer>> res = new ArrayList<>();
backtrack(res, candidates, target, new ArrayList<>(), 0);
return res;
}
void backtrack(
List<List<Integer>> res, int[] candidates, int target, List<Integer> list, int start) {
if (target == 0) {
res.add(new ArrayList<>(list));
return;
} else if (target > 0) {
for (int i = start; i < candidates.length; i++) {
if (target - candidates[i] < 0) {
break;
}
list.add(candidates[i]);
backtrack(res, candidates, target - candidates[i], list, i);
list.remove(list.size() - 1);
}
}
}
}
What to improve
- analyse more carefully