Update 075._sort_colors.md

docs/Leetcode_Solutions/075._sort_colors.md

@@ -31,7 +31,7 @@ Could you come up with a one-pass algorithm using only constant space?
 
 > 思路 1 
 
-先算一下0, 1, 2分别有多少个，然后in-place改呗，简单
+先算一下0, 1, 2分别有多少个，然后in-place改呗，简单, beats 100%
 
 ```python
 class Solution(object):
@@ -95,3 +95,55 @@ class Solution(object):
         		end -= 1
 ```
 
+> 思路 3
+
+两个指针也可以
+
+```python
+class Solution(object):
+    def sortColors(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: void Do not return anything, modify nums in-place instead.
+        """
+        i, l, r = 0, 0, len(nums) - 1
+        while i < len(nums):
+            if nums[i] == 2 and i < r:
+                nums[i], nums[r] = nums[r], 2
+                r -= 1
+            elif nums[i] == 0 and i > l:
+                nums[i], nums[l] = nums[l], 0
+                l += 1
+            else:
+                i += 1
+```
+
+> 思路 4
+
+这个方法就很巧妙了，我们遍历整个数组，只要碰到了什么数字我们就把这个数字往右边推一下
+
+大家可以用例子[2,0,2,1,1,0]自己推导一下过程，看看是不是有一种向右推的感觉
+
+
+```python
+class Solution(object):
+    def sortColors(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: void Do not return anything, modify nums in-place instead.
+        """
+        n0, n1, n2 = -1, -1, -1
+        for i in range(len(nums)):
+            if nums[i] == 0:
+                n0, n1, n2 = n0+1, n1+1, n2+1
+                nums[n2] = 2
+                nums[n1] = 1
+                nums[n0] = 0
+            elif nums[i] == 1:
+                n1, n2 = n1+1, n2+1
+                nums[n2] = 2
+                nums[n1] = 1
+            else:
+                n2 += 1
+                nums[n2] = 2
+```