LeetCode216.组合总数III

代码1(自己)

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> path = new ArrayList<>();
    public List<List<Integer>> combinationSum3(int k, int n) {
        backtracking(k, n, 1);
        return res;
    }
    public void backtracking(int k, int n, int startIndex) {
        if (path.size() == k) {
            if (getSum(path) == n) {
                res.add(new ArrayList<>(path)); 
                //return;         
            }
            return;
        }
        for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i ++) {
            path.add(i);
            backtracking(k, n, i + 1);
            path.remove(path.size() - 1);
        }
    }
    public int getSum(List<Integer> path) {
        int sum = 0;
        for (int i = 0; i < path.size(); i ++) {
            sum += path.get(i);
        }
        return sum; 
    }
}

剪枝操作： for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i ++) {}

代码2(代码随想录)

class Solution {
    List<List<Integer>> res = new ArrayList<>();
    List<Integer> path = new ArrayList<>();

    public List<List<Integer>> combinationSum3(int k, int n) {
        backtracking(k, n, 0, 1);
        return res;
    }
    public void backtracking(int k, int n, int sum, int startIndex) {
        if (path.size() == k) {
            if (sum == n) {
               res.add(new ArrayList<>(path)); 
            }
            return;
        }
        for (int i = startIndex; i <= 9 - (k - path.size()) + 1; i ++) {
            sum += i; // 处理
            path.add(i); // 处理
            backtracking(k, n, sum, i + 1); // 回溯
            sum -= i; // 撤销
            path.remove(path.size() - 1); // 撤销
        }
    }
}

注意：

恢复现场

处理过程和回溯过程是一一对应的，处理过程中有加法操作，回溯过程就要有对应的减法操作。