40. Combination Sum II 凑一个和，不能重复用一个元素

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]
]

踢掉自身就行了，为啥要踢掉相同的另一个元素？

递归的时候跳过相同的元素，在recursion中不能重复，避免实在是太多的嫌疑？
for循环却又保证了每个元素都能被利用。这样无形之中踢掉了自身。在for循环中重复

Here is the order that the elements get added to the list
[]
[1]
[1,2]
[1,2,3]
[1,3]理解一下回溯法：退回去一步，然后走到最深
[2]
[2,3]
[3]

 

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

private void backtrack(List<List<Integer>> list, List<Integer> tempList, int [] nums, int remain, int start){
    if(remain < 0) return;
    else if(remain == 0) list.add(new ArrayList<>(tempList));
    else{ 
        for(int i = start; i < nums.length; i++){
            if (i > start && (nums[i] == nums[i - 1])) continue;
            
            tempList.add(nums[i]);
            backtrack(list, tempList, nums, remain - nums[i], i + 1);
            tempList.remove(tempList.size() - 1);
        }
    }
}
}