剑指第二版/剑指Offer16.数值的整数次方.html

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    <title>剑指 Offer 16. 数值的整数次方</title>
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    <script>
      // https://leetcode-cn.com/problems/shu-zhi-de-zheng-shu-ci-fang-lcof/

      // 实现 pow(x, n) ，即计算 x 的 n 次幂函数（即，xn）。不得使用库函数，同时不需要考虑大数问题。
      // 提示：
      // -100.0 < x < 100.0
      // -231 <= n <= 231-1
      // -104 <= xn <= 104

      /**
       * @param {number} x
       * @param {number} n
       * @return {number}
       */
      var myPow = function (x, n) {};

      // --- answer-1 ---

      // 快速幂
      var myPow = function (x, n) {
        if (n === 0) return 1;
        if (n < 0) {
          x = 1 / x;
          n = -n;
        }
        let res = 1;
        while (n > 0) {
          if (n & 1) {
            res *= x;
          }
          // 这里使用>>越界了 2147483648 >> 1 === -1073741824
          n =  Math.floor(n / 2) ;
          x = x * x;
        }
        return res;
      };

      // --- answer-1 ---

      // --- answer-2 ---

      // --- answer-2 ---

      var x = 2.0,
        n = 10;
      var result = 1024.0;

      var x = 2.1,
        n = 3;
      var result = 9.261;

      var x = 2.0,
        n = -2;
      var result = 0.25;
      // 解释：2-2 = 1/22 = 1/4 = 0.25

      var x = 2.0,
        n = -2147483648;
      var result = 0;

      console.log("x = ", x);
      console.log("n = ", n);
      console.log("result = ", result);
      console.log("myPow = ", myPow(x, n));
    </script>
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