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Circular Array Loop.java
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Circular Array Loop.java
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/*
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?
*/
class Solution {
public boolean circularArrayLoop(int[] nums) {
if (nums == null) {
return false;
}
for (int i = 0; i < nums.length; i++) {
if (nums[i] != 0 && dfs(nums, i, 0) != 0) {
return true;
}
}
return false;
}
private int dfs(int[] nums, int index, int count) {
if (count >= nums.length) {
return nums[index];
}
int newIndex = (index + nums[index] + nums.length) % nums.length;
if (newIndex == index || nums[newIndex] * nums[index] < 0 || dfs(nums, newIndex, count + 1) == 0) {
nums[index] = 0;
}
return nums[index];
}
}
// 迭代
class Solution(object):
def circularArrayLoop(self, nums):
"""
:type nums: List[int]
:rtype: bool
"""
size = len(nums)
next = lambda x : (x + nums[x] + size) % size
for x in range(size):
if not nums[x]:
continue
y, c = x, 0
while c < size:
z = next(y)
if y == z:
nums[y] = 0
if nums[y] * nums[z] <= 0:
break
y = z
c += 1
if y == x:
return True
if c == size:
return True
y = x
while c > 0:
z = next(y)
nums[y] = 0
c -= 1
return False