LeetCode-Remove Invalid Parentheses
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: [""]
class Solution { public List<String> removeInvalidParentheses(String s) { Set<String> visited = new HashSet<>(); List<String> res = new ArrayList<>(); List<String> current = new ArrayList<>(); List<String> next; boolean reached = false; current.add(s); while(!current.isEmpty()){ next = new ArrayList<>(); for(String prv : current){ visited.add(prv); //if valid if(isValid(prv)){ reached = true; res.add(prv); } //not valid if(!reached){ for(int i=0; i<prv.length(); i++){ char tmp = prv.charAt(i); if(tmp == '(' || tmp == ')'){ String newString = prv.substring(0,i)+ prv.substring(i+1); if(!visited.contains(newString)){ next.add(newString); visited.add(newString); } } } } } if(reached){ break; } current = next; } return res; } private boolean isValid(String s){ Stack<Character> stack = new Stack<>(); for(int i=0; i<s.length(); i++){ char c = s.charAt(i); if(c == '('){ stack.push(c); } else if (c == ')'){ if(stack.isEmpty()){ return false; } else{ stack.pop(); } } } return stack.isEmpty(); } }
看到parenthese的问题,第一反应是用栈。这题要求minimum number,所以想到用BFS遍历解空间树。
思路为:
层次依次为删除0个元素,1个元素,2个元素。。。
层次遍历所有的可能。如果有一种可能是valid,那么不再遍历下面的层。
posted on 2018-07-02 06:08 IncredibleThings 阅读(136) 评论(0) 编辑 收藏 举报