40. Combination Sum II

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LeetCode

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Problem

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]
]

Constraints:

• 1 <= candidates.length <= 100
• 1 <= candidates[i] <= 50
• 1 <= target <= 30

Code

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(candidates);
        helper(res, candidates, target, new ArrayList<>(), 0);
        return res;
    }

    private void helper(List<List<Integer>> res,
                        int[] nums,
                        int target,
                        List<Integer> temp,
                        int start){
        if(target < 0) return;
        if(target == 0){
            res.add(new ArrayList<>(temp));
            return;
        }

        for(int i = start; i < nums.length; i++){
            // 重复的数字不能再使用了
            if(i != start && nums[i] == nums[i - 1]) continue;
            temp.add(nums[i]);
            helper(res, nums, target - nums[i], temp, i + 1);
            temp.remove(temp.size() - 1);
        }
    }
}

Time complexity: O(2^N) - 每个元素只能用一次, 每个元素只有两种选择，所以是二叉树

Space complexity: O(N) - 最多只会有 N 个元素在 temp 中; 这种计算并没有考虑 result 的空间