leetcode-22-Generate Parentheses

1.说明：

    题目要求找出n个括号所能组成的所有合法括号序列，所谓合法括号序列，指的是从左到右遍历序列，任意位置都满足，左括号的个数大于等于右括号的个数。

 2.代码：

/**
 * @param {number} n
 * @return {string[]}
 */
var generateParenthesis = function(n) {
    var stack = [];
    var result = [];
    var curLeft = 0;
    var curRight = 0;
    function getEnd(n) {
        var result = ''
        for (var i = 0; i < n; i++) {
            result += '()';
        }
        return result;
    }
    var end = getEnd(n);
    function reachEnd() {
        return stack.join('') == end;
    }
    function init() {
        for (var i = 0; i < n; i++) {
            stack.push('(')
        }
        for (var i = 0; i < n; i++) {
            stack.push(')')
        }
        result.push(stack.join(''));
    }
    init();
    while (!reachEnd()) {
        // 出栈
        while (true) {
            var ele = stack.pop();
            if (ele == '(') {
                curLeft++
                if (curLeft < curRight) {
                    break;
                }
            }
            else {
                curRight++
            }
        }
        // 入栈
        while (curRight) {
            stack.push(')');
            curRight--;
            while (curLeft) {
                stack.push('(')
                curLeft--
            }
            while (curRight) {
                stack.push(')')
                curRight--
            }
        }
        result.push(stack.join(''))
    }
    return result
};