来源:力扣(LeetCode)
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 1:
输入:n = 3
输出:["((()))","(()())","(())()","()(())","()()()"]
示例 2:
输入:n = 1
输出:["()"]
提示:
1 <= n <= 8
使用回溯法,分成3种情况:
class Solution {
private static final String[] parenthesis = {"(", ")"};
public List<String> generateParenthesis(int n) {
List<String> ans = new ArrayList<>();
Map<String, Integer> parenthesisTimes = new HashMap<>();
for (String ele: parenthesis) {
parenthesisTimes.put(ele, n);
}
backtrack(ans, parenthesisTimes, new StringBuilder());
return ans;
}
private void backtrack(List<String> ans, Map<String, Integer> remainParenthesis, StringBuilder sb) {
if (remainParenthesis.get("(") == 0 && remainParenthesis.get(")") == 0) {
ans.add(sb.toString());
return;
}
for (String ele: parenthesis) {
if (remainParenthesis.get("(").intValue() == remainParenthesis.get(")").intValue()) {
backtrackHelper(ans, sb, remainParenthesis, "(");
break;
} else if (remainParenthesis.get("(") > 0) {
backtrackHelper(ans, sb, remainParenthesis, ele);
} else {
backtrackHelper(ans, sb, remainParenthesis, ")");
break;
}
}
}
private void backtrackHelper(List<String> ans, StringBuilder sb, Map<String, Integer> remainParenthesis,
String ele) {
sb.append(ele);
remainParenthesis.put(ele, remainParenthesis.get(ele) - 1);
backtrack(ans, remainParenthesis, sb);
sb.deleteCharAt(sb.length() - 1);
remainParenthesis.put(ele, remainParenthesis.get(ele) + 1);
}
}
考虑种类问题时,首先使用回溯算法,要明确怎么枝剪