题解 | 小苯的IDE括号问题（easy）

const rl = require("readline").createInterface({ input: process.stdin });
var iter = rl[Symbol.asyncIterator]();
const readline = async () => (await iter.next()).value;
// 不是很复杂，但是本题的关键在于处理时间复杂度，直接使用原生API会导致时间超时，需要用指针方式替换，将时间复杂度降到 O(1)
void (async function () {
    // Write your code here
    let params = [];
    while ((line = await readline())) {
        let tokens = line.split(" ");
        // let a = parseInt(tokens[0]);
        // let b = parseInt(tokens[1]);
        // console.log(a + b);
        params.push(tokens);
    }
    // console.log(params)
    let [n, k] = params[0];
    let str_arr = params[1][0].split("");
    let left = [];
    let right = [];
    // let pos = str_arr.findIndex((v) => v == "I");
    // for (let i = 0; i < pos; i++) {
    //     left.push(str_arr[i]);
    // }
    // for (let j = pos + 1; j < str_arr.length; j++) {
    //     right.push(str_arr[j]);
    // }
    let i = 0;
    for (i = 0; i < str_arr.length; i++) {
        if (str_arr[i] != "I") {
            left.push(str_arr[i]);
        } else {
            break;
        }
    }
    for (let j = i + 1; j < str_arr.length; j++) {
        right.push(str_arr[j]);
    }
    // 使用指针而不是result 操作
    let left_pointer = left.length;
    let right_pointer = 0;
    // console.log(left, right);
    for (let i = 0; i < k; i++) {
        // console.log(params[i+2][0])
        let oper = params[i + 2][0];
        if (oper == "backspace") {
            if (
                left_pointer > 0 &&
                left[left_pointer - 1] == "(" &&
                right_pointer < right.length &&
                right[right_pointer] == ")"
            ) {
                // right.shift();
                left_pointer--;
                right_pointer++
            } else if (left_pointer > 0) {
                // left.pop();
                left_pointer--
            }
        } else if (oper == "delete") {
            if (right_pointer<right.length) {
                //   console.log('delete right')
                // right.shift();
                right_pointer++
            }
        }
    }
    console.log([...left.slice(0,left_pointer),'I',...right.slice(right_pointer)].join(""));
})();