solve 190.reverse-bits

andavid · andavid · commit 3a59bd21bfbe · 2019-04-20T23:06:47.000+08:00
diff --git a/vscode/190.reverse-bits.java b/vscode/190.reverse-bits.java
@@ -0,0 +1,69 @@
+/*
+ * @lc app=leetcode id=190 lang=java
+ *
+ * [190] Reverse Bits
+ *
+ * https://leetcode.com/problems/reverse-bits/description/
+ *
+ * algorithms
+ * Easy (30.22%)
+ * Total Accepted:    176.4K
+ * Total Submissions: 572K
+ * Testcase Example:  '00000010100101000001111010011100'
+ *
+ * Reverse bits of a given 32 bits unsigned integer.
+ *
+ *
+ *
+ * Example 1:
+ *
+ *
+ * Input: 00000010100101000001111010011100
+ * Output: 00111001011110000010100101000000
+ * Explanation: The input binary string 00000010100101000001111010011100
+ * represents the unsigned integer 43261596, so return 964176192 which its
+ * binary representation is 00111001011110000010100101000000.
+ *
+ *
+ * Example 2:
+ *
+ *
+ * Input: 11111111111111111111111111111101
+ * Output: 10111111111111111111111111111111
+ * Explanation: The input binary string 11111111111111111111111111111101
+ * represents the unsigned integer 4294967293, so return 3221225471 which its
+ * binary representation is 10101111110010110010011101101001.
+ *
+ *
+ *
+ * Note:
+ *
+ *
+ * Note that in some languages such as Java, there is no unsigned integer type.
+ * In this case, both input and output will be given as signed integer type and
+ * should not affect your implementation, as the internal binary representation
+ * of the integer is the same whether it is signed or unsigned.
+ * In Java, the compiler represents the signed integers using 2's complement
+ * notation. Therefore, in Example 2 above the input represents the signed
+ * integer -3 and the output represents the signed integer -1073741825.
+ *
+ *
+ *
+ *
+ * Follow up:
+ *
+ * If this function is called many times, how would you optimize it?
+ *
+ */
+public class Solution {
+  // you need treat n as an unsigned value
+  public int reverseBits(int n) {
+    int result = 0;
+    for (int i = 0; i < 32; i++) {
+      result = (result << 1) + (n & 1);
+      n = n >>> 1;
+    }
+    return result;
+  }
+}
+
