solve 50.pow-x-n

Author: andavid

vscode/50.pow-x-n.java

@@ -0,0 +1,65 @@
+/*
+ * @lc app=leetcode id=50 lang=java
+ *
+ * [50] Pow(x, n)
+ *
+ * https://leetcode.com/problems/powx-n/description/
+ *
+ * algorithms
+ * Medium (27.87%)
+ * Likes:    916
+ * Dislikes: 2236
+ * Total Accepted:    343.5K
+ * Total Submissions: 1.2M
+ * Testcase Example:  '2.00000\n10'
+ *
+ * Implement pow(x, n), which calculates x raised to the power n (x^n).
+ *
+ * Example 1:
+ *
+ *
+ * Input: 2.00000, 10
+ * Output: 1024.00000
+ *
+ *
+ * Example 2:
+ *
+ *
+ * Input: 2.10000, 3
+ * Output: 9.26100
+ *
+ *
+ * Example 3:
+ *
+ *
+ * Input: 2.00000, -2
+ * Output: 0.25000
+ * Explanation: 2^-2 = 1/2^2 = 1/4 = 0.25
+ *
+ *
+ * Note:
+ *
+ *
+ * -100.0 < x < 100.0
+ * n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1]
+ *
+ *
+ */
+class Solution {
+  public double myPow(double x, int n) {
+    if (n == 0) {
+      return 1;
+    }
+    double t = myPow(x, n/2);
+    if (n % 2 != 0) {
+      if (n < 0) {
+        return t * t / x;
+      } else {
+        return t * t * x;
+      }
+    } else {
+      return t * t;
+    }
+  }
+}
+