1006.笨阶乘

链接：1006.笨阶乘
难度：Medium
标签：栈、数学、模拟
简介：给定一个整数 N，它返回 N 的笨阶乘。

题解 1 - typescript

• 编辑时间：2021-04-01
• 执行用时：112ms
• 内存消耗：45.8MB
• 编程语言：typescript
• 解法介绍：先算出模板，再利用模板进行运算。

const op = ['*', '/', '+', '-'];
const opMap: Record<string, (num1: number, num2: number) => number> = {
  '*': (num1, num2) => num1 * num2,
  '/': (num1, num2) => ~~(num1 / num2),
  '+': (num1, num2) => num1 + num2,
  '-': (num1, num2) => num1 - num2,
};
function clumsy(N: number): number {
  let curOp = 0;
  const template: (number | string)[] = [];
  while (N) {
    template.push(N--);
    template.push(op[curOp]);
    curOp = (curOp + 1) % 4;
  }
  template.pop();
  const stack: (number | string)[] = [];
  for (let i = 0, l = template.length; i < l; i++) {
    const c = template[i];
    if (c === '*' || c === '/') {
      const num1 = stack.pop() as number;
      const num2 = template[++i] as number;
      stack.push(opMap[c](num1, num2));
    } else stack.push(c);
  }
  template.length = 0;
  template.push(...stack);
  stack.length = 0;
  for (let i = 0, l = template.length; i < l; i++) {
    const c = template[i];
    if (c === '+' || c === '-') {
      const num1 = stack.pop() as number;
      const num2 = template[++i] as number;
      stack.push(opMap[c](num1, num2));
    } else stack.push(c);
  }
  return stack[0] as number;
}