1--打印字符串的全部子序列
题目:
返回一个字符串的全部子序列(包括空串);
主要思路:
暴力枚举是否加入当前字符;
#include <iostream>
#include <vector>
#include <string>
class Solution{
public:
std::vector<std::string> Printstr(std::string str){
if(str.length() == 0) return res;
process(str, 0, "");
return res;
}
void process(std::string str, int i, std::string cur_str){
if(i == str.length()){
res.push_back(cur_str);
return;
}
process(str, i+1, cur_str+str[i]); // 加入当前字符
process(str, i+1, cur_str); // 不加入当前字符
}
private:
std::vector<std::string> res;
};
int main(int argc, char *argv[]){
Solution S1;
std::string test = "abc";
std::vector<std::string> res = S1.Printstr(test);
for(std::string str : res) std::cout << str << std::endl;
}
2--字符串的全排列
主要思路:
利用全排列遍历字符串,注意重复字符在树层上的剪枝(但要保留重复字符在树枝上的排列)
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
class Solution {
public:
std::vector<std::string> permutation(std::string s){
if(s.length() == 0) return res;
std::vector<bool> used(s.length(), false);
std::sort(s.begin(), s.end());
std::string tmp = "";
process(s, 0, tmp, used);
return res;
}
void process(std::string str, int idx, std::string tmp, std::vector<bool> used){
if(idx == str.length()){
res.push_back(tmp);
return;
}
for(int i = 0; i < str.length(); i++){
if(used[i] == true) continue; // 该字符已使用
// 去除树层重复字符
if(i > 0 && str[i] == str[i-1] && used[i-1] == true) continue;
used[i] = true;
process(str, idx + 1, tmp + str[i], used);
// 回溯
used[i] = false;
}
}
private:
std::vector<std::string> res;
};
int main(int argc, char *argv[]){
std::string test = "aab";
Solution S1;
std::vector<std::string> res = S1.permutation(test);
for(std::string str : res) std::cout << str << std::endl;
return 0;
}