题目描述:
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]]
解题思路:
本题因为数组的元素个数的个数没有限制,所以循环的就不便于实现,我用了递归的方法遍历添加1-9,直到元素的个数为k-1,比较最后一个数(n-数组中所有数)是否小于10且大于数组中的最后一个数。
代码:
1 class Solution { 2 public: 3 void select(vector<vector<int>> &ret, vector<int> nums, int k, int n, int index, int count){ 4 nums.push_back(index+1); 5 int len = nums.size(); 6 if(count == k-1){ 7 //这里我默认k的数值至少为2 8 for(int i = 0; i < len; i++){ 9 n -= nums[i]; 10 } 11 if(n < 10 && nums[len-1] < n){ 12 nums.push_back(n); 13 ret.push_back(nums); 14 } 15 } 16 for(int i = index+1; i < 9 && i < n; i++){ 17 //遍历向下添加 18 select(ret, nums, k, n, i, count+1); 19 } 20 } 21 vector<vector<int>> combinationSum3(int k, int n) { 22 vector<vector<int>> ret; 23 for(int i = 0; i < 8; i++) 24 select(ret, vector<int>() , k, n, i, 1); 25 return ret; 26 } 27 };