剑指第二版/剑指Offer10-II.青蛙跳台阶问题.html

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    <title>剑指 Offer 10- II. 青蛙跳台阶问题</title>
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    <script>
      // https://leetcode-cn.com/problems/qing-wa-tiao-tai-jie-wen-ti-lcof/

      // 一只青蛙一次可以跳上1级台阶，也可以跳上2级台阶。求该青蛙跳上一个 n 级的台阶总共有多少种跳法。
      // 答案需要取模 1e9+7（1000000007），如计算初始结果为：1000000008，请返回 1。
      // 提示：
      // 0 <= n <= 100

      /**
       * @param {number} n
       * @return {number}
       */

      var numWays = function (n) {};

      // --- answer-1 ---

      // 动态规划的 f(n) = f(n-1) + f(n-2) 即最后一次跳1个台阶和第二次跳两个台阶
      // n = 1 f(n) = 1 ; n = 2 f(n) = 2
      // 所以该题和菲波那切数列解法一致 只是初始值不同
      var numWays = function (n) {
        if (n === 0) return 1;
        if (n <= 2) return n;
        let pre = 1;
        let cur = 2;
        let res = 0;
        for (let i = 0; i < n - 2; i++) {
          res = (pre + cur) % 1000000007;
          pre = cur;
          cur = res;
        }
        return res;
      };

      // --- answer-1 ---

      // --- answer-2 ---

      // --- answer-2 ---

      // var n = 2;
      // var result = 2;

      var n = 7;
      var result = 21;

      // var n = 0;
      // var result = 1;

      console.log('n = ', n);
      console.log('result = ', result);
      console.log('numWays = ', numWays(n));
    </script>
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