40. Combination Sum II (Medium)
https://leetcode.com/problems/combination-sum-ii/
Given a collection of candidate numbers (candidates
) and a target number (target
), find all unique combinations in candidates
where the candidate numbers sums to target
.
Each number in candidates
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.
Example 1:
Input: candidates =[10,1,2,7,6,1,5]
, target =8
, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
Example 2:
Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ]
Solutions
ppublic class Solution {
public List<List<Integer>> combinationSum2(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;
}
for (int i = depth; i < nums.length; i++) {
// optimization, no need to go deeper
if (nums[i] > target) {
break;
}
// Since nums[depth] is not compared with previous one nums[depth-1].
// For combination [1,1...] will still be included and if not add following statement
// there will be a problem of duplicate answers.
if (i != depth && nums[i] == nums[i - 1]) {
continue;
}
solution.add(nums[i]);
// Since every element in nums can be chosen only once
// logic should start from i+1, not i
combine(ans, solution, nums, target - nums[i], i + 1);
solution.remove(solution.size() - 1);
}
}
}