Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:

• Input: head = [1,2,3,4,5]
• Output: [5,4,3,2,1]

Example 2:

• Input: head = [1,2]
• Output: [2,1]

Constraints:

• 0 ≤ list length ≤ 5000

Reverse Linked List is the foundational linked list problem testing pointer manipulation.

Pattern: Pointer Reversal

Use three pointers: previous, current, and next. Save the next node before breaking the link.

graph LR
    Head["ListNode (Head)"] --> P1["Pointer 1 (curr / slow)"]
    Head --> P2["Pointer 2 (prev / fast)"]
    P1 -->|Iterate / Reverse| Next["Next Node"]
    P2 -->|Traverse 2x| Next
    Next --> Verdict["Return Modified Head / Result"]

// Iterative is optimal

• Time Complexity: O(n)
• Space Complexity: O(1)
• Explanation: The standard iterative approach.

function reverseList(head) {
  let prev = null, curr = head;
  while (curr !== null) {
    const nextTemp = curr.next;
    curr.next = prev;
    prev = curr;
    curr = nextTemp;
  }
  return prev;
}

• Time Complexity: O(n)
• Space Complexity: O(1)
• Explanation: Three-pointer iterative reversal.

1. Initialize State: Setup necessary pointers, dynamic programming arrays, or hash maps.
2. Iterate & Evaluate: Process the input according to the boundary conditions.
3. Update & Return: Compute the optimal answer and return early or at termination.

1. Draw a simple list and show pointer reversal
2. Explain prev, curr, nextTemp
3. Walk through one iteration

1. Use three pointers: prev, curr, nextTemp.
2. Save the next node before redirecting.
3. Return prev as the new head.

👉 Solve this problem interactively in the DSA Lab