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:
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
一刷
題解：就是在dfs里面使用兩個變量leftCount和rightCount來決定如何加括號卦洽。 注意在加括號前要判斷l(xiāng)eftCount和rightCount是否大于0苍蔬， 并且之后要注意回溯。
public class Solution {
public List<String> generateParenthesis(int n) {
List<String> res = new ArrayList<>();
if (n <= 0) return res;
generateParenthesis(res, new StringBuilder(), n, n);
return res;
}
private void generateParenthesis(List<String> res, StringBuilder sb, int leftCount, int rightCount) {
if (rightCount < leftCount) return;
if (leftCount == 0 && rightCount == 0) res.add(sb.toString());
if (leftCount > 0) {
generateParenthesis(res, sb.append("("), leftCount - 1, rightCount);
sb.setLength(sb.length() - 1);
}
if (rightCount > 0) {
generateParenthesis(res, sb.append(")"), leftCount, rightCount - 1);
sb.setLength(sb.length() - 1);
}
}
}
二刷
dfs, 注意條件
public class Solution {
public List<String> generateParenthesis(int n) {
List<String> res = new ArrayList<>();
StringBuilder sb = new StringBuilder();
generateP(res, sb, 0, 0, n);
return res;
}
private void generateP(List<String> res, StringBuilder sb, int left, int right, int n){
if(left == right && left == n){
res.add(sb.toString());
return;
}
if(left<n){
sb.append("(");
generateP(res, sb, left+1, right, n);
sb.setLength(sb.length()-1);
}
if(right<n && right<left){
sb.append(")");
generateP(res, sb, left, right+1, n);
sb.setLength(sb.length()-1);
}
}
}
