LeetCode題301—Remove Invalid Parentheses
阿新 • • 發佈:2018-09-18
gen res bool all nim set集合 () div mov
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 )
.
Example 1:
Input: "()())()" Output: ["()()()", "(())()"]
Example 2:
Input: "(a)())()" Output: ["(a)()()", "(a())()"]
Example 3:
Input: ")(" Output: [""]
思路:可以利用DFS或者BFS來解這道題,感覺還是BFS簡單點,即對於從給定的字符串通過移除 ( 或 ) 來構造所有可能的合法串,如果合法就加入到set集合中,不合法到到下一輪的BFS中。
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();
// 如果當前層次中有合法解的話,只需要將當前層次中的字符串全部彈出判斷是否合法,停止BFS,這樣保證所得到的合法字符串是移除最少字符得到的
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;
}
}
LeetCode題301—Remove Invalid Parentheses