22. Generate Parentheses
- 方法1: backtracking
-
- 易錯點
- 易錯點
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:
用遞歸來生成下一個字元,那麼我們需要排除一些不合法的順序。我們用left和right來傳遞可用的左右括号數量。如果left > right 說明遞歸進來的字元串有先于左括号出現的右括号,不予以處理直接傳回。如果left <= right,但是left = right = 0,此時已經産生一個合了解,需要推入結果。否則left剩下可以推left,right剩下可以推right,将相應的可用數量-1。
易錯點
- 這裡傳遞tmp的時候不能是&,因為tmp + “(” is not lvalue。那麼要麼選擇pass by value,要麼再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;
}
};
另一種寫法,判斷的條件也稍微發生改變:如果left < n就可以繼續放入左括号,但是右括号必須滿足right > left 才允許進行推入(可以加快不少),并且用push/pop來操作tmp。
易錯點
- 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;
}
};