[Vssue]0216.组合总和III.md

[youngyangyang04]

https://www.programmercarl.com/0216.%E7%BB%84%E5%90%88%E6%80%BB%E5%92%8CIII.html

[guguroro]

把sum的剪枝操作放在for循环判断条件里也可以

[LiWangTian]

提供下我的代码，sum在递归中算每次sum-i class Solution { List<List> result = new ArrayList<>(); List path = new ArrayList<>(); public List<List> combinationSum3(int k, int n) { backtacking(k,n,1); return result; }

private void backtacking(int k,int n, int startIndex){
    if(path.size() == k && n==0){
        result.add(new ArrayList<>(path));
        return;
    }
    for(int i = startIndex;i<=9;i++){
        path.add(i);
        backtacking(k,n-i,i+1);
        path.remove(path.size()-1);
    }
}

} //剪枝操作 class Solution { List<List> result = new ArrayList<>(); List path = new ArrayList<>(); public List<List> combinationSum3(int k, int n) { backtacking(k,n,1); return result; }

private void backtacking(int k,int n, int startIndex){
    if(path.size() == k && n==0){
        result.add(new ArrayList<>(path));
        return;
    }
    for(int i = startIndex;i<=9;i++){
        path.add(i);
        backtacking(k,n-i,i+1);
        path.remove(path.size()-1);
    }
}

}

[Du1in9]

private void backtracking(int n, int k, int sum, int start) {
    if (sum > n) return; 
    if (path.size() == k) {
        if (sum == n) {
            paths.add(new ArrayList<>(path));
        }
        return;
    }
    for (int i = start; i <= 9 - (k - path.size()) + 1; i++) {
        sum += i;
        path.add(i);
        backtracking(n, k, sum, i + 1);
        sum -= i;
        path.remove(path.size() - 1);
    }
}

// 深度1: backtracking(9, 3, 0, 1), 在 1 ~ 7 中取一个数, path = []
    i = 1：path = [1]，递归 backtracking(9, 3, 0, 2)，回溯，path = [] ...
    i = 7：path = [7]，递归 backtracking(9, 3, 0, 8)，回溯，path = []

// 深度2: backtracking(9, 3, 0, 2~8), 在 (2~8) ~ 8 中取一个数, path = [1~7]
// 例: backtracking(9, 3, 0, 2): 在 2 ~ 8 中取一个数, path = [1]
    i = 2：path = [1,2]，递归 backtracking(9, 3, 0, 3)，回溯，path = [1] ...
    i = 8：path = [1,8]，递归 backtracking(9, 3, 0, 9)，回溯，path = [1]

// 深度3: backtracking(9, 3, 0, 3~9), 在 (3~9) ~ 9 中取一个数, path = [1~7,2~8]
// 例: backtracking(9, 3, 0, 3): 在 3 ~ 9 中取一个数, path = [1,2]
    i = 3：path = [1,2,3]，长度为 3，判断 sum 并处理，回溯，path = [1,2] ...
    i = 9：path = [1,2,9]，长度为 3，判断 sum 并处理，回溯，path = [1,2]

[tjuzc]

加油

[stcirclear]

Go版本

func combinationSum3(k int, n int) [][]int {
    res := make([][]int, 0)
    arr := make([]int, 0)
    sum := 0
    var backtrace func(start int) 
    backtrace = func(start int) {
        // 找到和为n的k个数的组合
        if len(arr) == k && sum == n {
            // 避免arr底层数组的改变对结果造成的影响，所以copy一份
            tmp := make([]int, k)
            copy(tmp, arr)
            res = append(res, tmp)
        } 

        for i := start; i <= 9; i++ {
            // 剪枝
            // 再选一个数怎么也会超过n || 剩下的数个数不能满足了
            if sum + i > n || 9 - i + 1 < k - len(arr) {
                break
            }
            // 做选择
            arr = append(arr, i)
            sum += i
            // 选下一个数
            backtrace(i + 1)
            // 撤销选择
            sum -= i
            arr = arr[:len(arr)-1]
        }
    }
    // 从1开始
    backtrace(1)
    return res
}