Update 457._Circular_Array_Loop.md

docs/Leetcode_Solutions/457._Circular_Array_Loop.md

@@ -28,7 +28,7 @@ Can you do it in O(n) time complexity and O(1) space complexity?
 > 思路 1
 ******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
 
-快慢指针，
+快慢指针，然后要注意的是如果loop只有一个元素不算loop，每次找完一圈后发现没有找到的话要将这一路上经过的点全都设为0防止下次重复查找了
 
 
 ```python
@@ -46,23 +46,25 @@ class Solution(object):
             if val == 0:
                 continue
 
-            j = i
-            k = get_next_idx(i)
+            slow = i
+            fast = get_next_idx(i)
 
-            while nums[k] * val > 0 and nums[get_next_idx(k)] * val > 0:
-                if j == k:
-                    if j == get_next_idx(j):
+            while nums[fast] * val > 0 and nums[get_next_idx(fast)] * val > 0:
+                if slow == fast:
+                    # exclude for loop with only one element
+                    if slow == get_next_idx(slow):
                         break
                     return True
 
-                j = get_next_idx(j)
-                k = get_next_idx(get_next_idx(k))
+                slow = get_next_idx(slow)
+                fast = get_next_idx(get_next_idx(fast))
 
-            j = i
-            while nums[j] * val > 0:
-                nxt = get_next_idx(j)
-                nums[j] = 0
-                j = nxt
+            slow = i
+            # set all the element along the path to 0, avoiding repeated lookup
+            while nums[slow] * val > 0:
+                nxt = get_next_idx(slow)
+                nums[slow] = 0
+                slow = nxt
 
         return False
 ```