22. Generate Parentheses
- 方法1: backtracking
-
- 易错点
- 易错点
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
For example, given n = 3, a solution set is:
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
方法1: backtracking
思路:
用递归来生成下一个字符,那么我们需要排除一些不合法的顺序。我们用left和right来传递可用的左右括号数量。如果left > right 说明递归进来的字符串有先于左括号出现的右括号,不予以处理直接返回。如果left <= right,但是left = right = 0,此时已经产生一个合理解,需要推入结果。否则left剩下可以推left,right剩下可以推right,将相应的可用数量-1。
易错点
- 这里传递tmp的时候不能是&,因为tmp + “(” is not lvalue。那么要么选择pass by value,要么再push/pop操作tmp
class Solution1 {
public:
vector<string> generateParenthesis(int n) {
vector<string> result;
string tmp = "";
generateHelper(result, tmp, n, n);
return result;
}
void generateHelper(vector<string> & result, string tmp, int left, int right){
if (left > right){
return;
}
if (left == 0 && right == 0){
result.push_back(tmp);
}
if (left > 0){
generateHelper(result, tmp + "(", left - 1, right);
}
if (right > 0){
generateHelper(result, tmp + ")", left, right - 1);
}
return;
}
};
另一种写法,判断的条件也稍微发生改变:如果left < n就可以继续放入左括号,但是右括号必须满足right > left 才允许进行推入(可以加快不少),并且用push/pop来操作tmp。
易错点
- right > left 而不是right >= left
class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> result;
string tmp = "";
generateHelper(result, tmp, n, n);
return result;
}
void generateHelper(vector<string> & result, string & tmp, int left, int right){
if (left == 0 && right == 0){
result.push_back(tmp);
}
if (left > 0){
tmp.push_back('(');
generateHelper(result, tmp, left - 1, right);
tmp.pop_back();
}
if (right > left){
tmp.push_back(')');
generateHelper(result, tmp, left, right - 1);
tmp.pop_back();
}
return;
}
};