[Function add]

1. Add leetcode solutions.

Author: Seanforfun

README.md

@@ -303,6 +303,12 @@
 
 [168. Excel Sheet Column Title](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/168.%20Excel%20Sheet%20Column%20Title.md)
 
+[169. Majority Element](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/169.%20Majority%20Element.md)
+
+[171. Excel Sheet Column Number](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/171.%20Excel%20Sheet%20Column%20Number.md)
+
+[172. Factorial Trailing Zeroes](https://github.com/Seanforfun/Algorithm-and-Leetcode/blob/master/leetcode/172.%20Factorial%20Trailing%20Zeroes.md)
+
 ## Algorithm(4th_Edition)
 Reading notes of book Algorithm(4th Algorithm),ISBN: 9787115293800.
 All java realization codes are placed in different packages.

leetcode/169. Majority Element.md

@@ -1,4 +1,22 @@
 ## 169. Majority Element
+
+### Question:
+Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
+
+You may assume that the array is non-empty and the majority element always exist in the array.
+
+```
+Example 1:
+
+Input: [3,2,3]
+Output: 3
+
+Example 2:
+
+Input: [2,2,1,1,1,2,2]
+Output: 2
+```
+
 ### Thinking:
 * Method1:通过hashmap实现，速度较慢
 
@@ -54,4 +72,25 @@ class Solution {
         return major;
     }
 }
+```
+
+### 二刷
+还是记得一刷时候的方法三。
+```Java
+class Solution {
+    public int majorityElement(int[] nums) {
+        int count = 1;
+        int save = nums[0];
+        for(int i = 1; i < nums.length; i++){
+            if(count == 0){
+                ++count;
+                save = nums[i];
+                continue;
+            }
+            if(nums[i] == save) count++;
+            else count--;
+        }
+        return save;
+    }
+}
 ```

leetcode/171. Excel Sheet Column Number.md

@@ -0,0 +1,71 @@
+## 171. Excel Sheet Column Number
+
+### Question:
+Given a column title as appear in an Excel sheet, return its corresponding column number.
+
+```
+For example:
+
+    A -> 1
+    B -> 2
+    C -> 3
+    ...
+    Z -> 26
+    AA -> 27
+    AB -> 28 
+    ...
+
+Example 1:
+
+Input: "A"
+Output: 1
+
+Example 2:
+
+Input: "AB"
+Output: 28
+
+Example 3:
+
+Input: "ZY"
+Output: 701
+```
+
+### Thinking:
+* Method1:通过hashmap实现，速度较慢
+
+```Java
+class Solution {
+    public int titleToNumber(String s) {
+        Map<Character, Integer> map = new HashMap<>();
+        Character c = 'Z';
+        for(int i = 26; i > 0; i--){
+            map.put(c, i);
+            c--;
+        }
+        int count = 0;
+        if(s.length() == 0) return count;
+        for(int i = 0; i < s.length(); i++){
+            count = map.get(s.charAt(i)) + count * 26;
+        }
+        return count;
+    }
+}
+```
+
+### 二刷
+1. 一刷中用hashmap唯一的用处就是记录char和int的对应关系，但是我们完全可以通过当前char的数值和A之间的差值+1获得数字。
+```Java
+class Solution {
+    public int titleToNumber(String s) {
+        if(s == null || s.length() == 0) return 0;
+        int result = 0;
+        for(int i = 0; i < s.length(); i++){
+            result *= 26;
+            int c = s.charAt(i) - 'A' + 1;
+            result += c;
+        }
+        return result;
+    }
+}
+```

leetcode/172. Factorial Trailing Zeroes.md

@@ -35,3 +35,18 @@ class Solution {
     }
 }
 ```
+
+### 二刷
+![Imgur](https://i.imgur.com/IBejjyd.jpg)
+```Java
+class Solution {
+    public int trailingZeroes(int n) {
+        int result = 0;
+        while(n > 4){
+            result += n / 5;
+            n /= 5;
+        }
+        return result;
+    }
+}
+```