[LeetCode]Circular Array Loop

题目描述：

LeetCode 457. Circular Array Loop

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.

Example 3: Given the array [2, 0, 2, 1, 3], return false since 0 is not supposed to appear in the array.

Can you do it in O(n) time complexity and O(1) space complexity?

题目大意：

给定一个整数数组。如果某下标位置的数字n为正数，则向前移动n步。反之，如果是负数，则向后移动-n步。假设数组首尾相接。判断数组中是否存在环。环中至少包含2个元素。环中的元素一律“向前”或者一律“向后”。

满足O(n)时间复杂度和O(1)空间复杂度。

解题思路：

三次遍历数组：

第一次遍历，将所有指向自己的元素置0

第二次遍历，从0到n（循环一周），将指向非负数的负数置0

第三次遍历，从n到0（循环一周），将指向非正数的正数置0

遍历结束后，如果数组中存在非零元素，则返回True；否则返回False

Python代码：

class Solution(object):
    def circularArrayLoop(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        if not nums or not all(nums): return False
        size = len(nums)
        next = lambda x : (x + size + nums[x]) % size
        for x in range(size):
            if next(x) == x:
                nums[x] = 0
        for x in range(size + 1):
            x = x % size
            if nums[x] < 0 and nums[next(x)] >= 0:
                nums[x] = 0
        for x in range(size, -1, -1):
            x = x % size
            if nums[x] > 0 and nums[next(x)] <= 0:
                nums[x] = 0
        return any(nums)