Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results.
Note: The input string may contain letters other than the parentheses ( and ).
Examples:
"()())()" -> ["()()()", "(())()"]
"(a)())()" -> ["(a)()()", "(a())()"]
")(" -> [""]
Solution：dfs recursive
思路：
https://discuss.leetcode.com/topic/34875/easy-short-concise-and-fast-java-dfs-3-ms-solution
Time Complexity: O() Space Complexity: O()
Solution Code：
class Solution {
public List<String> removeInvalidParentheses(String s) {
List<String> ans = new ArrayList<>();
remove(s, ans, 0, 0, new char[]{'(', ')'});
return ans;
}
public void remove(String s, List<String> ans, int last_i, int last_j, char[] par) {
// for wrong ')'
for (int stack = 0, i = last_i; i < s.length(); ++i) {
if (s.charAt(i) == par[0]) stack++;
if (s.charAt(i) == par[1]) stack--;
if (stack >= 0) continue;
for (int j = last_j; j <= i; ++j)
if (s.charAt(j) == par[1] && (j == last_j || s.charAt(j - 1) != par[1]))
remove(s.substring(0, j) + s.substring(j + 1, s.length()), ans, i, j, par);
return;
}
// for wrong '('
String reversed = new StringBuilder(s).reverse().toString();
if (par[0] == '(') // finished left to right
remove(reversed, ans, 0, 0, new char[]{')', '('});
else {// finished right to left
ans.add(reversed);
}
}
}
