457. Circular Array Loop
You are given a circular array nums
of positive and negative integers. If a number k at an index is positive, then move forward k steps. Conversely, if it's negative (- k ), move backward k steps. Since the array is circular, you may assume that the last element's next element is the first element, and the first element's previous element is the last element.
Determine if there is a loop (or a cycle) in nums
. A cycle must start and end at the same index and the cycle's length > 1. Furthermore, movements in a cycle must all follow a single direction. In other words, a cycle must not consist of both forward and backward movements.
Example 1:
Input: [2,-1,1,2,2]
Output: true
Explanation: There is a cycle, from index 0 -> 2 -> 3 -> 0. The cycle's length is 3.
Example 2:
Input: [-1,2]
Output: false
Explanation: The movement from index 1 -> 1 -> 1 ... is not a cycle, because the cycle's length is 1. By definition the cycle's length must be greater than 1.
Example 3:
Input: [-2,1,-1,-2,-2]
Output: false
Explanation: The movement from index 1 -> 2 -> 1 -> ... is not a cycle, because movement from index 1 -> 2 is a forward movement, but movement from index 2 -> 1 is a backward movement. All movements in a cycle must follow a single direction.
Note:
- -1000 ≤ nums[i] ≤ 1000
- nums[i] ≠ 0
- 1 ≤ nums.length ≤ 5000
Follow up:
Could you solve it in O(n) time complexity and O(1) extra space complexity?
# @lc code=start
using LeetCode
function circular_array_loop(nums::Vector{Int})
len = length(nums)
for i in 1:length(nums)
if nums[i] == -2000 || nums[i] == 2000
continue
end
j = mod1(i + nums[i], len)
lst = i
nums[i] = 2000 * sign(nums[i])
while true
if nums[i] * nums[j] < 0 || j == lst
break
elseif abs(nums[j]) == 2000 && j ≥ i
return true
end
nums[j], j, lst = 2000 * sign(nums[j]), mod1(nums[j] + j, len), j
end
end
return false
end
# @lc code=end
circular_array_loop (generic function with 1 method)
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