From bd19f160a9dc33894732f24451dc19c2aa6ea2ff Mon Sep 17 00:00:00 2001
From: initc <1875919175@qq.com>
Date: Sat, 27 Feb 2016 13:09:00 +0800
Subject: [PATCH] add SingleNumberTwo --find three time number except one

---
 bin/.gitignore                   |  1 +
 src/com/jie/SingleNumberTwo.java | 30 ++++++++++++++++++++++++++++++
 2 files changed, 31 insertions(+)
 create mode 100644 bin/.gitignore
 create mode 100644 src/com/jie/SingleNumberTwo.java

diff --git a/bin/.gitignore b/bin/.gitignore
new file mode 100644
index 0000000..c2d9872
--- /dev/null
+++ b/bin/.gitignore
@@ -0,0 +1 @@
+/com/
diff --git a/src/com/jie/SingleNumberTwo.java b/src/com/jie/SingleNumberTwo.java
new file mode 100644
index 0000000..16a9df1
--- /dev/null
+++ b/src/com/jie/SingleNumberTwo.java
@@ -0,0 +1,30 @@
+package com.jie;
+
+public class SingleNumberTwo {
+
+	public static void main(String[] args) {
+		// TODO Auto-generated method stub
+
+	}
+
+	 public int singleNumber(int[] nums) {
+	        int [] count = new int[32];
+			int result=0;
+			for(int i=0;i<32;i++){
+				for(int j=0;j<nums.length;j++){
+					//循环区得每一位之和   使用无符号向右移
+					count[i]+=(nums[j]>>>i)&1;
+				}
+				//获取那一个只出现了一次的数  因为除了一个每一个数都出现了三次  所以我们出现的数的
+				//和肯定是三的倍数   最后一个没有出现的数  要不然是0要不然是1  所以%3即可
+				count[i]=count[i]%3==0?0:1;
+			}
+			//构造最后的结果
+			for(int i=0;i<32;i++){
+				
+				result|=count[i]<<i;
+			}
+			return result;
+	    }
+	
+}