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。 注意在加括號前要判斷l(xiāng)eftCount和rightCount是否大于0， 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);
        }
    }
}