Create 457._Circular_Array_Loop.md

docs/Leetcode_Solutions/457._Circular_Array_Loop.md

+# 457. Circular Array Loop
+
+**<font color=red>难度: Medium</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode.com/problems/circular-array-loop/description/
+
+> 内容描述
+
+```
+
+You are given an array of positive and negative integers. If a number n at an index is positive, then move forward n steps. Conversely, if it's negative (-n), move backward n steps. Assume the first element of the array is forward next to the last element, and the last element is backward next to the first element. Determine if there is a loop in this array. A loop starts and ends at a particular index with more than 1 element along the loop. The loop must be "forward" or "backward'.
+
+Example 1: Given the array [2, -1, 1, 2, 2], there is a loop, from index 0 -> 2 -> 3 -> 0.
+
+Example 2: Given the array [-1, 2], there is no loop.
+
+Note: The given array is guaranteed to contain no element "0".
+
+Can you do it in O(n) time complexity and O(1) space complexity?
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
+
+快慢指针，
+
+
+```python
+class Solution(object):
+    def circularArrayLoop(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: bool
+        """
+        def get_next_idx(i):
+            n = len(nums)
+            return (i + nums[i] + n) % n
+
+        for i, val in enumerate(nums):
+            if val == 0:
+                continue
+
+            j = i
+            k = 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):
+                        break
+                    return True
+
+                j = get_next_idx(j)
+                k = get_next_idx(get_next_idx(k))
+
+            j = i
+            while nums[j] * val > 0:
+                nxt = get_next_idx(j)
+                nums[j] = 0
+                j = nxt
+
+        return False
+```