Remove Invalid Parentheses
題意:
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())()"] ")(" -> [""]
解題思路:
使用bfs去作,一次刪掉一個再下去作遞迴,注意下面判斷valid的函式,不需要使用stack,直接用一個count值即可,一旦找到了之後,只要把同一層的產生完即可,不需再往下bfs了,程式碼如下:
public class Solution {
public List<String> removeInvalidParentheses(String s) {
List<String> res = new ArrayList<>();
// sanity check
if (s == null) return res;
Set<String> visited = new HashSet<>();
Queue<String> queue = new LinkedList<>();
// initialize
queue.add(s);
visited.add(s);
boolean found = false;
while (!queue.isEmpty()) {
s = queue.poll();
if (isValid(s)) {
// found an answer, add to the result
res.add(s);
found = true;
}
if (found) continue;
// generate all possible states
for (int i = 0; i < s.length(); i++) {
// we only try to remove left or right paren
if (s.charAt(i) != '(' && s.charAt(i) != ')') continue;
String t = s.substring(0, i) + s.substring(i + 1);
if (!visited.contains(t)) {
// for each state, if it's not visited, add it to the queue
queue.add(t);
visited.add(t);
}
}
}
return res;
}
// helper function checks if string s contains valid parantheses
boolean isValid(String s) {
int count = 0;
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if (c == '(') count++;
if (c == ')' && count-- == 0) return false;
}
return count == 0;
}
}