39. Combination Sum (Medium)
https://leetcode.com/problems/combination-sum/
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:
Input: candidates =[2,3,6,7],target =7, A solution set is: [ [7], [2,2,3] ]
Example 2:
Input: candidates = [2,3,5], target = 8,
A solution set is:
[
[2,2,2,2],
[2,3,3],
[3,5]
]
Solutions
public class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> ans = new ArrayList<>();
if (candidates == null) {
return ans;
}
Arrays.sort(candidates);
combine(ans, new ArrayList<>(), candidates, target, 0);
return ans;
}
private void combine(List<List<Integer>> ans, List<Integer> solution, int[] nums, int target, int depth) {
if (target < 0) {
return;
}
if (target == 0) {
ans.add(new ArrayList<>(solution));
return;
}
// i starts from depth, it also can start from 0, but will produce duplicates
// The reason why we start from depth is that (1, 0) is equivalent with (0, 1)
for (int i = depth; i < nums.length; i++) {
// optimization, no need to go deeper
if (nums[i] > target) {
break;
}
// If the the nums contains duplicate values, skip it
// However, this problem guarantees no duplicates
if (i > depth && nums[i] == nums[i - 1]) {
continue;
}
solution.add(nums[i]);
combine(ans, solution, nums, target - nums[i], i);
solution.remove(solution.size() - 1);
}
}
}