40. Combination Sum II

40. 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]
]

My Solutions:

和combination sum类似，但因为有duplicates，多加一个去掉duplicate的条件，

if (i > start && nums[i] == nums[i - 1]) continue;

是为了保证第一个数不被跳掉，同时不会有相同结果数组出现。

比如[1,1,2,2], target=5, 结果只可能有[1,2,2], 这里的1是candidates里的第一个1，不会是第二个1。

注意dfs里dfs到下一次是i+1，不是上题的i， 因为一个数字只能加一次

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> list = new ArrayList<>();
        Arrays.sort(candidates); // 不可省略
        dfs(list, new ArrayList<Integer>(), candidates, target, 0);
        return list;
    }
    
    public void dfs(List<List<Integer>> list, List<Integer> temp, int[] nums, int remain, int start) {
        if (remain < 0) return;
        if (remain == 0) list.add(new ArrayList<Integer>(temp)); // 满足条件直接返回
        else {
            for (int i = start; i < nums.length; i++){
                if (i > start && nums[i] == nums[i - 1]) continue; // 跳过重复数字
                temp.add(nums[i]);
                dfs(list, temp, nums, remain - nums[i], i + 1); // 这里用下一个数字
                temp.remove(temp.size() - 1);
            }

        }
    }
}