【LeetCode】40. Combined summation II (2 strokes of wrong questions)

• Ideas
  Each number can only be used once in each combination , The same number may appear more than once .
  The difficulty of this problem lies in removing the weight , The same number can appear on the same layer , But the same number can no longer appear in the same place , Such as [1, 1, 1, 7], The goal is 8, It's supposed to be [[1, 7]], Instead of going back, there will be [[1,7], [1,7], [1,7]], Therefore, it is necessary to mark whether the same number on the same layer has been used .
  

func sum(arr []int) int {
    
    s := 0
    for _, v := range arr {
    
        s += v
    }
    return s
}

func combinationSum2(candidates []int, target int) [][]int {
    
    res := [][]int{
    }
    arr := []int{
    }
    // used[i] == true, Indicates that the same branch has been used 
    // used[i] == false, Indicates that the same layer has been used 
    used := make([]bool, len(candidates))
    var backtracking func([]int, int, int, []bool)
    backtracking = func(candidates []int, target int, start int, used []bool) {
    
        if sum(arr) == target {
    
            temp := make([]int, len(arr))
            copy(temp, arr)
            res = append(res, temp)
        }
        for i := start; i < len(candidates); i++ {
    
            //  Same layer weight removal 
            if i > 0 && candidates[i] == candidates[i - 1] && used[i - 1] == false {
    
                continue
            }
            if sum(arr) + candidates[i] > target {
    
                break
            }
            arr = append(arr, candidates[i])
            used[i] = true
            backtracking(candidates, target, i + 1, used)
            arr = arr[: len(arr) - 1]
            used[i] = false
        }
    }

    //  Sort 
    sort.Ints(candidates)
    backtracking(candidates, target, 0, used)
    return res
}