Given the head of a singly linked list, reverse the list, and return the reversed list. The key insight is to use three pointers: prev, curr, and next to iteratively reverse each link.

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Pythontime O(n) · space O(1)

def reverseList(head):
    prev, curr = None, head
    while curr:
        next_node = curr.next  # save next
        curr.next = prev       # reverse pointer
        prev = curr            # advance prev
        curr = next_node       # advance curr
    return prev

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