leetcode Combinations计算有多少组合

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Combinations

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

For example,
If n = 4 and k = 2, a solution is:

[  [2,4],  [3,4],  [2,3],  [1,2],  [1,3],  [1,4],]

典型的递归回溯法，多做形成这个递归回溯头脑，就好办了。

1 递归一次，填入一个数字

2 填入的数字，不能是小于当前数字的值，防止重复

3 回溯：记得pop_back()最后加上的一个数字，回溯到上一层。

4 结束条件：填写够了k个数字的时候，当前填写完毕，回溯

vector<vector<int> > combine(int n, int k) {vector<vector<int> > v;if (k > n || k==0) return v;vector<int> mv;comb(v, mv, n, k, 1);return v;}void comb(vector<vector<int> > &v, vector<int> &mv, int n, int k, int start){for (int i = start; i <= n; i++){mv.push_back(i);if (mv.size() == k){v.push_back(mv);mv.pop_back();continue;}//i<n条件可以不加，因为递归的时候也自动判断排除这个条件了if (i < n) comb(v, mv, n, k, i+1);mv.pop_back();}}

递归的另外一种写法：

void comb2(vector<vector<int> > &v, vector<int> &mv, int n, int k, int start){if (mv.size() == k){v.push_back(mv);return;}for (int i = start; i <= n; i++){mv.push_back(i);comb2(v, mv, n, k, i+1);mv.pop_back();}}

//2014-2-12 updatevector<vector<int> > combine(int n, int k) {vector<vector<int> > rs;if (k<1 || n<k) return rs;vector<int> tmp;helper(rs, tmp, n, k);return rs;}void helper(vector<vector<int> > &rs,vector<int> &tmp, int n, int k, int begin = 1){if (!k){rs.push_back(tmp);return;}for (int i = begin; i <= n; i++){tmp.push_back(i);helper(rs,tmp,n,k-1,i+1);tmp.pop_back();}}

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