LaiCode 179. All Valid Permutations Of Parentheses II

最新推荐文章于 2025-03-29 17:08:26 发布

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最新推荐文章于 2025-03-29 17:08:26 发布

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本文链接：https://blog.youkuaiyun.com/qq_31361913/article/details/107740106

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本文探讨了如何生成所有有效的括号组合，即所有符合括号规则的括号字符串。这些字符串必须满足开闭括号匹配，形成合法的括号表达式。通过递归或动态规划的方法，我们可以枚举所有可能的组合，并检查它们是否符合有效性条件。

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public class Solution {
  public List<String> validParentheses(int l, int m, int n) {
    // Write your solution here
    int targetLen = (l + m + n) * 2;
    StringBuilder curr = new StringBuilder();
    List<String> res = new ArrayList<>();
    Deque<Character> stack = new LinkedList<>();
    char[] ps = new char[] {'(', ')', '<', '>', '{', '}'};
    int[] remain = new int[] {l, l, m, m, n, n};
    helper(ps, remain, targetLen, stack, curr, res);
    return res;
  }
  private void helper(char[] ps, int[] remain, int targetLen, Deque<Character> stack, StringBuilder curr, List<String> res){
    if(curr.length() == targetLen){
      res.add(curr.toString());
      return;
    }
    for(int i = 0; i < ps.length; i++){
      // Pick up a left parenthesis
      if(i % 2 == 0){
        if(remain[i] > 0){
          curr.append(ps[i]);
          stack.offerFirst(ps[i]);
          remain[i]--;
          helper(ps, remain, targetLen, stack, curr, res);
          curr.deleteCharAt(curr.length() - 1);
          stack.pollFirst();
          remain[i]++;
        }
      }else{
      // Pick up a right parenthesis
        // Here we use isEmpty() method instead of empty() method
        if(!stack.isEmpty() && stack.peekFirst() == ps[i - 1]){
          curr.append(ps[i]);
          // Poll the left parenthesis
          stack.pollFirst();
          remain[i]--;
          helper(ps, remain, targetLen, stack, curr, res);
          curr.deleteCharAt(curr.length() - 1);
          // Push the left parenthesis
          stack.offerFirst(ps[i - 1]);
          remain[i]++;
        }
      }
    }
  }
}