标签：details lock leetcode backtrac 查看 har app end 组合
https://leetcode-cn.com/problems/generate-parentheses
数字 n 代表生成括号的对数，请你设计一个函数，用于能够生成所有可能的并且 有效的 括号组合。
示例：
输入：n = 3
输出：[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
vector<string> generateParenthesis(int n) {}
用递归生成所有的序列：
长度为 \(n-1\) 的序列后加上 ‘(‘ 或者 ‘)‘，就可以得到长度为 \(n\) 的序列。
检查序列是否有效：
设置一个变量 \(count\) 记录未匹配的左括号的数量，
遍历序列中的每一个元素 \(c\)：
?若 \(c\) 为 ‘(‘，\(count\) 值加一
?若 \(c\) 为 ‘)‘，\(count\) 值减一
??如果 \(count\) 的值小于零，则序列无效
最后，若 \(count\) 为零则序列有效，否则为无效
class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> result;
string current;
generate_all(current, n * 2, result);
return result;
}
void generate_all(string& current, int n, vector<string>& result){
if(n == current.size()){
if(valid(current)){
result.push_back(current);
}
return;
}
current.push_back(‘(‘);
generate_all(current, n, result);
current.pop_back();
current.push_back(‘)‘);
generate_all(current, n, result);
current.pop_back();
}
bool valid(string& s){
int count = 0;
for(char c : s){
if(c == ‘(‘){
count++;
}else{
count--;
if(count < 0){
return false;
}
}
}
return count == 0;
}
};
class Solution {
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList();
generateAll(new StringBuilder(), n * 2, result);
return result;
}
void generateAll(StringBuilder current, int n, List<String> result){
if(current.length()== n){
if(valid(current.toString())){
result.add(current.toString());
}
}else{
current.append(‘(‘);
generateAll(current, n, result);
current.deleteCharAt(current.length() - 1);
current.append(‘)‘);
generateAll(current, n, result);
current.deleteCharAt(current.length() - 1);
}
}
boolean valid(String current){
int count = 0;
for(int i = 0; i < current.length(); i++){
if(current.charAt(i) == ‘(‘){
count++;
}else{
count--;
if(count < 0){
return false;
}
}
}
return count == 0;
}
}
改进上面方法：在生成序列的过程中，根据当前情况只继续生成有效的序列
如果左括号的数量小于 \(n\) ，可以再添加一个左括号
如果右括号的数量小于左括号的数量，可以再添加一个右括号
时间复杂度：\(O(n^2)\)，空间复杂度：\(O(1)\)。
class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> result;
string current;
backTrack(result, current, 0, 0, n);
return result;
}
void backTrack(vector<string>& result, string& current, int left, int right, int n){
if(current.size() == n * 2){
result.push_back(current);
return;
}
if(left < n){
current.push_back(‘(‘);
backTrack(result, current, left + 1, right, n);
current.pop_back();
}
if(right < left){
current.push_back(‘)‘);
backTrack(result, current, left, right + 1, n);
current.pop_back();
}
}
};
class Solution {
public List<String> generateParenthesis(int n) {
List<String> result = new ArrayList<>();
backTrack(result, new StringBuilder(), 0, 0, n);
return result;
}
void backTrack(List<String> result, StringBuilder current, int left, int right, int n){
if(current.length() == n * 2){
result.add(current.toString());
return;
}
if(left < n){
current.append(‘(‘);
backTrack(result, current, left + 1, right, n);
current.deleteCharAt(current.length() - 1);
}
if(right < left){
current.append(‘)‘);
backTrack(result, current, left, right + 1, n);
current.deleteCharAt(current.length() - 1);
}
}
}
标签：details lock leetcode backtrac 查看 har app end 组合
原文地址：https://www.cnblogs.com/crazyBlogs/p/13164045.html
