Leetcode 216. Combination Sum III

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

s思路：
1. 题目没有说清楚，但看举例，应该是1-9的每个数只能取一次。
2. 枚举所有可能，套路的做法是backtrack:for+recursive.这道题数字只有1-9，集合很小，看如何剪枝了。一般backtrack都需要适当剪枝，所谓剪枝，就是提前预测这种组合没必要尝试下去，降低不必要的尝试的次数！下面的表达式就是剪枝用的，n<=0,不用干了，n/k>9或n/k<当前已经访问到的数，也都没必要干了！

if(n<=0||n/k>9||n/k<digit) return;

//方法1：backtracking:一次搞定！妙！class Solution {public:    void helper(vector<vector<int>>&res,vector<int>&cur,int k,int n,int digit){        //        if(k==0){            if(n==0) res.push_back(cur);                return;        }        if(n<=0||n/k>9||n/k<digit) return;//剪枝。不剪枝，速度3ms;剪枝，0ms!        for(int i=digit;i<=9;i++){            cur.push_back(i);            helper(res,cur,k-1,n-i,i+1);            cur.pop_back();         }       }    vector<vector<int>> combinationSum3(int k, int n) {        //        vector<vector<int>> res;        vector<int> cur;        helper(res,cur,k,n,1);        return res;    }};