From 93b6613f685e36a2d4210fe14f46399f76f483d0 Mon Sep 17 00:00:00 2001
From: wycgi520 <437891878@qq.com>
Date: Fri, 14 Dec 2018 15:55:19 +0800
Subject: [PATCH 1/2] modify

---
 ...81211-lengthOfLongestSubstring-Archer-1.py | 63 +++++++++++++++++++
 ...81214-lengthOfLongestSubstring-Archer-2.py | 46 ++++++++++++++
 ...ngestSubstring\350\257\264\346\230\216.md" | 16 ++++-
 Archer/readme.md                              |  3 +
 4 files changed, 127 insertions(+), 1 deletion(-)
 create mode 100644 Archer/lengthOfLongestSubstring/20181211-lengthOfLongestSubstring-Archer-1.py
 create mode 100644 Archer/lengthOfLongestSubstring/20181214-lengthOfLongestSubstring-Archer-2.py

diff --git a/Archer/lengthOfLongestSubstring/20181211-lengthOfLongestSubstring-Archer-1.py b/Archer/lengthOfLongestSubstring/20181211-lengthOfLongestSubstring-Archer-1.py
new file mode 100644
index 0000000..3986e2b
--- /dev/null
+++ b/Archer/lengthOfLongestSubstring/20181211-lengthOfLongestSubstring-Archer-1.py
@@ -0,0 +1,63 @@
+#!/usr/bin/env python
+# -*- coding:utf-8 -*-
+
+s = "vdlkdk"
+dict = {}
+num = 0
+out = 0
+flag = 0
+i = 0
+#暴力法1：找到所有不重复的字符串
+'''
+for i in range(0,len(s)) :
+    dict[s[i]] = i
+    num += 1
+    for j in range(i+1,len(s)) :
+        if s[j] in dict :
+            break
+        dict[s[j]] = j
+        num += 1
+    out = [num, out][out > num]
+    if out >= len(s) / 2 :
+        break
+    dict.clear()
+    num = 0
+'''
+
+#暴力法2：从输出长度out入手
+'''
+for out in range(0,len(s)):
+    for i in range(0,out+1):
+        for j in range(i,i+len(s)-out):
+            if s[j] in dict:
+                dict.clear()
+                break
+            dict[s[j]] = j
+        else:
+            flag = 1
+        if flag == 1:
+            break
+    if flag == 1:
+        break
+
+print len(s)-out
+'''
+#滑动法，当往右滑动到右重复字符时，从前面重复的那个字符后开始搜索，不断循环
+'''
+n = len(s)
+while i < n:
+    if s[i] in dict:
+        i = dict[s[i]] + 1
+        out = [num, out][out > num]
+        num = 0
+        dict.clear()
+        dict[s[i]] = i
+        num += 1
+        i += 1
+        continue
+    dict[s[i]] = i
+    num += 1
+    i += 1
+out = [num, out][out > num]
+'''
+
diff --git a/Archer/lengthOfLongestSubstring/20181214-lengthOfLongestSubstring-Archer-2.py b/Archer/lengthOfLongestSubstring/20181214-lengthOfLongestSubstring-Archer-2.py
new file mode 100644
index 0000000..62a4eed
--- /dev/null
+++ b/Archer/lengthOfLongestSubstring/20181214-lengthOfLongestSubstring-Archer-2.py
@@ -0,0 +1,46 @@
+#!/usr/bin/env python
+# -*- coding:utf-8 -*-
+
+
+s = ""
+set0 = set()
+dict = {}
+num = 0
+out = 0
+flag = 0
+i = 0
+j = 0
+
+n = len(s)
+
+
+# 索引i和j分别表示子字符串的起始和结束，如何有相同的字符，则一直把前面的删掉，知道没有重复为止
+'''
+while i<n and j<n:
+    if (s[j] in dict) == False :
+        dict[s[j]] = j
+        j += 1
+        out = [j-i,out][out>(j-i)]
+    else :
+        del dict[s[i]]
+        i += 1
+'''
+# 直接找字符串，用find，如果找到，先得到字符串大小，再把索引i往前移到找到的索引，优点是不用一直把前面的删掉，进一步优化效率
+
+j = 1
+i = 0
+
+
+while j<n :
+    cur = s.find(s[j],i,j)
+    if cur != -1:
+        out = max(j-i,out)
+        i = cur+1
+    j += 1
+
+if n != 0:
+    out = max(j-i,out)
+else :
+    out = 0
+
+print out
\ No newline at end of file
diff --git "a/Archer/lengthOfLongestSubstring/lengthOfLongestSubstring\350\257\264\346\230\216.md" "b/Archer/lengthOfLongestSubstring/lengthOfLongestSubstring\350\257\264\346\230\216.md"
index 71da795..f087e4e 100644
--- "a/Archer/lengthOfLongestSubstring/lengthOfLongestSubstring\350\257\264\346\230\216.md"
+++ "b/Archer/lengthOfLongestSubstring/lengthOfLongestSubstring\350\257\264\346\230\216.md"
@@ -10,4 +10,18 @@
 
 # 复盘 #
 
-应该还可以改进另一种算法，更简洁
\ No newline at end of file
+应该还可以改进另一种算法，更简洁
+
+# 最终总结 #
+
+该问题可以表述为：
+
+对于字符串s
+
+设索引i和j表示子字符串的起始和结尾，i=0，j=0
+
+在j不断增加的过程中，判断s[j]是否在i和j间的子字符串中(重复值在子字符串中的位置标记为cur)：
+
+若是，得出字符串长度j-i，取输出out和j-i的最大值更新out，并且移动索引i：
+
+i=cur+1(也可以一步一步移动i：i=i+1，不过这样的话j要在没有重复时才能往后增加)
diff --git a/Archer/readme.md b/Archer/readme.md
index 1c49ef1..0322a5d 100644
--- a/Archer/readme.md
+++ b/Archer/readme.md
@@ -1,3 +1,6 @@
+##1213-1215 打卡##
+**给定一个字符串，请你找出其中不含有重复字符的 最长子串 的长度。** 
+
 ##1210-1212 打卡##
 **给定一个字符串，请你找出其中不含有重复字符的 最长子串 的长度。** 
 

From 933eff890e25a3096a2c69cd9f6fb8dec38f014b Mon Sep 17 00:00:00 2001
From: wycgi520 <437891878@qq.com>
Date: Tue, 18 Dec 2018 15:51:48 +0800
Subject: [PATCH 2/2] modify

---
 Archer/Happy Number/Happy Number.py           | 28 +++++++++++++++++++
 ...16\345\217\212\345\244\215\347\233\230.md" |  9 ++++++
 Archer/readme.md                              |  3 ++
 Archer/twoSum/twoSum.py                       |  7 -----
 4 files changed, 40 insertions(+), 7 deletions(-)
 create mode 100644 Archer/Happy Number/Happy Number.py
 create mode 100644 "Archer/Happy Number/Happy Number\350\257\264\346\230\216\345\217\212\345\244\215\347\233\230.md"

diff --git a/Archer/Happy Number/Happy Number.py b/Archer/Happy Number/Happy Number.py
new file mode 100644
index 0000000..3e0be3d
--- /dev/null
+++ b/Archer/Happy Number/Happy Number.py	
@@ -0,0 +1,28 @@
+#!/usr/bin/env python
+# -*- coding:utf-8 -*-
+
+class Solution(object):
+    def isHappy(self, n):
+        """
+        :type n: int
+        :rtype: bool
+        """
+
+        dict_1 = {}
+
+        while 1:
+            sum = 0
+
+            while n != 0:  #计算平方之和
+                sum += (n % 10) ** 2
+                n /= 10
+
+            n = sum
+
+            if n == 1:   #如果平方之和为1，证明n是快乐数，返回True
+                return True
+            elif dict_1.has_key(n):  #若平方之和已出现过（循环计算），证明加法之和不可能为1，n不是快乐数，返回False
+                return False
+
+            dict_1[n] = 0
+
diff --git "a/Archer/Happy Number/Happy Number\350\257\264\346\230\216\345\217\212\345\244\215\347\233\230.md" "b/Archer/Happy Number/Happy Number\350\257\264\346\230\216\345\217\212\345\244\215\347\233\230.md"
new file mode 100644
index 0000000..05eb5af
--- /dev/null
+++ "b/Archer/Happy Number/Happy Number\350\257\264\346\230\216\345\217\212\345\244\215\347\233\230.md"	
@@ -0,0 +1,9 @@
+# 算法实现：(详细见代码文档) #
+
+#编写一个算法来判断一个数是不是“快乐数”。一个“快乐数”定义为：对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和，然后重复这个过程直到这个数变为 1，也可能是无限循环但始终变不到 1。如果可以变为 1，那么这个数就是快乐数。 #
+
+# 复盘 #
+
+1. 用两个字典，一个字典存储加法结果各位数的平方，另一个字典存储平方之和
+
+2. 用一个字典存储平方之和，判断有无循环即可，不需要存储各个平方，直接加在一起便是
\ No newline at end of file
diff --git a/Archer/readme.md b/Archer/readme.md
index 0322a5d..5bde0eb 100644
--- a/Archer/readme.md
+++ b/Archer/readme.md
@@ -1,3 +1,6 @@
+##1217-1219 打卡##
+**编写一个算法来判断一个数是不是“快乐数”。一个“快乐数”定义为：对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和，然后重复这个过程直到这个数变为 1，也可能是无限循环但始终变不到 1。如果可以变为 1，那么这个数就是快乐数。** 
+
 ##1213-1215 打卡##
 **给定一个字符串，请你找出其中不含有重复字符的 最长子串 的长度。** 
 
diff --git a/Archer/twoSum/twoSum.py b/Archer/twoSum/twoSum.py
index f7c49cf..062767c 100644
--- a/Archer/twoSum/twoSum.py
+++ b/Archer/twoSum/twoSum.py
@@ -50,13 +50,6 @@ def twoSum(nums, target):
     for i in range(0, k):
         temp = target - nums[i]
         if temp in map_a :
-            print map_a.items()
             return [map_a[temp], i]
         map_a[nums[i]] =  i
     '''
-
-a = [3, 3, 4, 5]
-
-b = twoSum(a,8)
-
-print b
\ No newline at end of file