22. Generate Parentheses

Problem:

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

Solution:

ONE

Recursive DFS backtracking.

/**
 * @param {number} n
 * @return {string[]}
 */
let generateParenthesis = function (n) {
  const result = [];
  if (n > 0) {
    dfs(n, 0, 0, "", result);
  }
  return result;
};

function dfs(n, nopen, nclose, path, result) {
  if (path.length === n * 2) {
    result.push(path);
    return;
  }

  if (nopen < n) {
    dfs(n, nopen + 1, nclose, path + "(", result);
  }

  if (nclose < nopen) {
    dfs(n, nopen, nclose + 1, path + ")", result);
  }
}

TWO

BFS.

/**
 * @param {number} n
 * @return {string[]}
 */
let generateParenthesis = function (n) {
  if (n <= 0) {
    return [];
  }

  const queue = [
    {
      path: "(",
      open: 1,
      close: 0,
    },
  ];

  while (true) {
    const { path, open, close } = queue.shift();
    if (open + close === n * 2) {
      queue.unshift({ path, open, close });
      break;
    }

    if (open < n) {
      queue.push({
        path: path + "(",
        open: open + 1,
        close,
      });
    }

    if (close < open) {
      queue.push({
        path: path + ")",
        open,
        close: close + 1,
      });
    }
  }

  return queue.map((x) => x.path);
};

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