From ff65ebea6fa9564c85bff0661629e4d306c0b6ab Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Tue, 23 Oct 2018 20:39:17 +0800
Subject: [PATCH] Create 191. Number of 1 Bits.md

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+# [191. Number of 1 Bits](https://leetcode.com/problems/number-of-1-bits/description/)
+# 思路
+## 思路一
+不断循环，每次循环得到最低位的值，最后可得到所有位1的个数。怎样得到最低位的值，两种方法:
+* 1、若num为奇数，则num的最低位肯定为1; 
+* 2、用一个只有低位是1的mask与num进行与操作即可得到最低位；
+## 思路二*
+依然是循环，但是每次循环不是得到最低位的值，而是每次循环去掉一个1，用一个count计数即可得到答案。
+# C++
+## 思路一
+```
+// 方法1
+class Solution {
+public:
+    int hammingWeight(uint32_t n) {
+        int res = 0;
+        while(n != 0){
+            res += (n % 2);
+            n /= 2;
+        }
+        return res;
+    }
+};
+
+// 方法2
+class Solution {
+public:
+    int hammingWeight(uint32_t n) {
+        int res = 0;
+        uint32_t mask = 1;
+        for(int i = 0; i < 32; i++){
+            res += (n & mask);
+            n = n >> 1;
+        }
+        return res;
+    }
+};
+```
+## 思路二*
+```
+class Solution {
+public:
+    int hammingWeight(uint32_t n) {
+        int count = 0;
+        while (n) {
+            n &= (n - 1); // 去掉最后的1
+            count++;
+        }
+        return count;
+    }
+};
+```