删除最小数量的无效括号,使得输入的字符串有效,返回所有可能的结果。
说明: 输入可能包含了除 ( 和 ) 以外的字符。
示例 1:
输入: “()())()”
输出: ["()()()", “(())()”]
示例 2:
输入: “(a)())()”
输出: ["(a)()()", “(a())()”]
示例 3:
输入: “)(”
输出: [""]
class Solution {
public List removeInvalidParentheses(String s) {
int left = 0, right = 0;
char[] cs = s.toCharArray();
for(char c : cs) {
if(c == '(') {
left++;
}else if(c == ')') {
if(left == 0) right++;
else left--;
}
}
List res = new ArrayList();
backtrace(cs, 0, new StringBuilder(s.length()-left-right), res, 0, 0, left, right);
return res;
}
private void backtrace(char[] cs, int cur, StringBuilder sb, List res,
int left, int right, int remL, int remR) {
if(cur == cs.length) {
if(remL == 0 && remR == 0) res.add(sb.toString());
return;
}
if(right > left) return;
final int len = sb.length();
if(cs[cur] == '(') {
// use
sb.append('(');
backtrace(cs, cur+1, sb, res, left+1, right, remL, remR);
sb.setLength(len);
if(remL > 0) { // not use
while(cur = 0) backtrace(cs, cur, sb, res, left, right, remL, remR);
}
}else if(cs[cur] == ')') {
// use
sb.append(')');
backtrace(cs, cur+1, sb, res, left, right+1, remL, remR);
sb.setLength(len);
if(remR > 0) { // not use
while(cur = 0) backtrace(cs, cur, sb, res, left, right, remL, remR);
}
}else {
sb.append(cs[cur]);
backtrace(cs, cur+1, sb, res, left, right, remL, remR);
sb.setLength(len);
}
}
}