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C++算法思维+算法模板
result = []
def backtrack(路径, 选择列表):
if 满足结束条件:
result.add(路径)
return
for 选择 in 选择列表:
做选择
backtrack(路径, 选择列表)
撤销选择
LeetCode 22. Generate Parentheses https://leetcode-cn.com/problems/generate-parentheses/
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
Example 1:
Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
Example 2:
Input: n = 1
Output: ["()"]
Constraints:
1 <= n <= 8
void backtrack(int n, int i, string& track) {
// i 代表当前的位置，共 2n 个位置
// 穷举到最后一个位置了，得到一个长度为 2n 组合
if (i == 2 * n) {
print(track);
return;
}
// 对于每个位置可以是左括号或者右括号两种选择
for choice in ['(', ')'] {
track.push(choice); // 做选择
// 穷举下一个位置
backtrack(n, i + 1, track);
track.pop(choice); // 撤销选择
}
}
See more -> 题解模板.md
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