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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:
"((()))", "(()())", "(())()", "()(())", "()()()"
Java Solution 1 – DFS
This solution is simple and clear. In the dfs() method, left stands for the remaining number of (, right stands for the remaining number of ).
public List<String> generateParenthesis(int n) {
ArrayList<String> result = new ArrayList<String>();
dfs(result, "", n, n);
return result;
}
/*
left and right represents the remaining number of ( and ) that need to be added.
When left > right, there are more ")" placed than "(". Such cases are wrong and the method stops.
*/
public void dfs(ArrayList<String> result, String s, int left, int right){
if(left > right)
return;
if(left==0&&right==0){
result.add(s);
return;
}
if(left>0){
dfs(result, s+"(", left-1, right);
}
if(right>0){
dfs(result, s+")", left, right-1);
}
}
Generate Parentheses Java Solution 2
This solution looks more complicated. You can use n=2 to walk through the code.
public List<String> generateParenthesis(int n) {
ArrayList<String> result = new ArrayList<String>();
ArrayList<Integer> diff = new ArrayList<Integer>();
result.add("");
diff.add(0);
for (int i = 0; i < 2 * n; i++) {
ArrayList<String> temp1 = new ArrayList<String>();
ArrayList<Integer> temp2 = new ArrayList<Integer>();
for (int j = 0; j < result.size(); j++) {
String s = result.get(j);
int k = diff.get(j);
if (i < 2 * n - 1) {
temp1.add(s + "(");
temp2.add(k + 1);
}
if (k > 0 && i < 2 * n - 1 || k == 1 && i == 2 * n - 1) {
temp1.add(s + ")");
temp2.add(k - 1);
}
}
result = new ArrayList<String>(temp1);
diff = new ArrayList<Integer>(temp2);
}
return result;
}
Object-oriented programming offers a sustainable way to write spaghetti code. It lets you accrete programs as a series of patches.”
― Paul Graham, Hackers & Painters: Big Ideas from the Computer Age
You can also check out Longest Valid Parentheses