diff --git a/Algorithms/0191. Number of 1 Bits/191. Number of 1 Bits.go b/Algorithms/0191. Number of 1 Bits/191. Number of 1 Bits.go
index 56afa32b..94f2697d 100644
--- a/Algorithms/0191. Number of 1 Bits/191. Number of 1 Bits.go	
+++ b/Algorithms/0191. Number of 1 Bits/191. Number of 1 Bits.go	
@@ -1,6 +1,14 @@
 package leetcode
 
+import "math/bits"
+
+// 解法一
 func hammingWeight(num uint32) int {
+	return bits.OnesCount(uint(num))
+}
+
+// 解法二
+func hammingWeight1(num uint32) int {
 	count := 0
 	for num != 0 {
 		num = num & (num - 1)
diff --git a/Algorithms/0191. Number of 1 Bits/README.md b/Algorithms/0191. Number of 1 Bits/README.md
index c160d6e2..b7a435d3 100755
--- a/Algorithms/0191. Number of 1 Bits/README.md	
+++ b/Algorithms/0191. Number of 1 Bits/README.md	
@@ -36,4 +36,4 @@ Write a function that takes an unsigned integer and return the number of '1' b
 
 - 求 uint32 数的二进制位中 1 的个数。
 - 这一题的解题思路就是利用二进制位操作。`X = X & ( X -1 )` 这个操作可以清除最低位的二进制位 1，利用这个操作，直至把数清零。操作了几次即为有几个二进制位 1 。
-
+- 最简单的方法即是直接调用库函数 `bits.OnesCount(uint(num))` 。