Combinations

?Total Accepted:?10949?Total Submissions:?36507My Submissions

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], ]

解法1。2： DFS遍歷全部組合。注意推斷當前cur的長度以及當前能夠值的狀態能否夠繼續
解法3，4：用next（prev） permutation來實現獲取全部組合
解法5： 類似非遞歸的DFS（或者能夠想象成為整數逐步加1進位，遍歷全部可能數值）來編譯狀態
解法6：長度從1到k，每次把全部長度為i- 1的已有組合遍歷一次。取全部可能的組合形成長度為i的全部組合（題目要求從小到大。這也為我們這樣實現提供的根據）



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;} };

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