解法1。2： DFS遍历全部组合。注意推断当前cur的长度以及当前能够值的状态能否够继续
解法3，4：用next（prev） permutation来实现获取全部组合
解法5： 类似非递归的DFS（或者能够想象成为整数逐步加1进位，遍历全部可能数值）来编译状态
解法6：长度从1到k，每次把全部长度为i- 1的已有组合遍历一次。取全部可能的组合形成长度为i的全部组合（题目要求从小到大。这也为我们这样实现提供的根据）

class Solution {
public:vector<vector<int> > combine(int n, int k) {// IMPORTANT: Please reset any member data you declared, as// the same Solution instance will be reused for each test case.vector<vector<int>> ret;if (k < 1 || k > n) return ret;// 1/*vector<int> cur;genCombination1(ret, cur, n, k, 1);*/// 2/*vector<int> cur;genCombination2(ret, cur, n, k, 1);*/// 3/*vector<int> flag(n, 0);for (int i = 0; i < k; ++i) {flag[i] = 1;}do {ret.push_back(vector<int>());auto it = back_inserter(ret.back());for (int i = 0; i < n; ++i) {if (flag[i] == 1) it = i + 1;}//} while (prev_permutation(flag.begin(), flag.end()));} while (m_prev_permutation(flag));*/// 4/*vector<int> flag(n, 0);for (int i = n - k; i < n; ++i) {flag[i] = 1;}do {ret.push_back(vector<int>());auto it = back_inserter(ret.back());for (int i = 0; i < n; ++i) {if (flag[i] == 1) it = i + 1;}//} while (next_permutation(flag.begin(), flag.end()));} while (m_next_permutation(flag));*/// 5/*vector<int> lo(k, 0), hi(k, 0);for (int i = 1; i <= k; ++i) {lo[i - 1] = i;hi[i - 1] = n - k + i;}while (true) {ret.push_back(lo);int cpos = k - 1;while (cpos >= 0) {if (lo[cpos] < hi[cpos]) {++lo[cpos];for (int i = cpos + 1; i < k; ++i) {lo[i] = lo[i - 1] + 1;}break;}--cpos;}if (cpos < 0) break;}*/// 6for (int i = 1; i <= n - k + 1; ++i) {ret.push_back(vector<int>(1, i));}for (int i = 2; i <= k; ++i) {int cnt = ret.size();for (int j = 0; j < cnt; ++j) {int cval = ret[j].back();int lastval = n - k + i;for (int t = cval + 1; t < lastval; ++t) {ret.push_back(ret[j]);ret.back().push_back(t);}ret[j].push_back(lastval);}}return ret;}private:void genCombination1(vector<vector<int>> &ret, vector<int> &cur, int n, int k, int cval) {if (n - cval + 1 < k - cur.size()) return;if (cur.size() == k) {ret.push_back(cur);return;}cur.push_back(cval);genCombination1(ret, cur, n, k, cval + 1);cur.pop_back();genCombination1(ret, cur, n, k, cval + 1);}void genCombination2(vector<vector<int>> &ret, vector<int> &cur, int n, int k, int cval) {int cnt = cur.size();if (cnt == k) {ret.push_back(cur);return;}for (int i = cval; i <= n - k + cnt + 1; ++i) {cur.push_back(i);genCombination2(ret, cur, n, k, i + 1);cur.pop_back();}}bool m_prev_permutation(vector<int> &f) {for (int i = f.size() - 1; i > 0; --i) {if (f[i] >= f[i - 1]) continue;int spos = i;for (int j = spos + 1; j < f.size(); ++j) {if (f[j] >= f[spos] && f[j] < f[i - 1])spos = j;}swap(f[spos], f[i - 1]);reverse(f.begin() + i, f.end());return true;}return false;}bool m_next_permutation(vector<int> &f) {for (int i = f.size() - 1; i > 0; --i) {if (f[i] <= f[i - 1]) continue;int spos = i;for (int j = spos + 1; j < f.size(); ++j) {if (f[j] > f[i - 1] && f[j] <= f[spos])spos = j;}swap(f[spos], f[i - 1]);reverse(f.begin() + i, f.end());return true;}return false;}
};