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9.5 Write a method to compute all permutations of a string.

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LeetCode上的原題，請參加我之前的博客Permutations 全排列和Permutations II 全排列之二。

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解法一：

class Solution { public:vector<string> getPerms(string &s) {vector<string> res;getPermsDFS(s, 0, res);return res;}void getPermsDFS(string &s, int level, vector<string> &res) {if (level == s.size()) res.push_back(s);else {for (int i = level; i < s.size(); ++i) {swap(s[level], s[i]);getPermsDFS(s, level + 1 ,res);swap(s[level], s[i]);}}} };

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解法二：

class Solution { public:vector<string> getPerms(string s) {vector<string> res;vector<bool> visited(s.size(), false);getPermsDFS(s, 0, visited, "", res);return res;}void getPermsDFS(string s, int level, vector<bool> &visited, string out, vector<string> &res) {if (level == s.size()) res.push_back(out);else {for (int i = 0; i < s.size(); ++i) {if (!visited[i]) {visited[i] = true;out.push_back(s[i]);getPermsDFS(s, level + 1, visited, out , res);out.pop_back();visited[i] = false;} }}} };

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解法三：

class Solution { public:vector<string> getPerms(string s) {if (s.empty()) return vector<string>(1, "");vector<string> res;char first = s[0];string remainder = s.substr(1);vector<string> words = getPerms(remainder);for (auto &a : words) {for (int i = 0; i <= a.size(); ++i) {string out = insertCharAt(a, first, i);res.push_back(out);}}return res;} string insertCharAt(string s, char c, int i) {string start = s.substr(0, i);string end = s.substr(i);return start + c + end;} };

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轉(zhuǎn)載于:https://www.cnblogs.com/grandyang/p/4822859.html

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