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Formatted question description: https://leetcode.ca/all/1290.html
1290. Convert Binary Number in a Linked List to Integer (Easy)
Given head
which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number.
Return the decimal value of the number in the linked list.
Example 1:
Input: head = [1,0,1] Output: 5 Explanation: (101) in base 2 = (5) in base 10
Example 2:
Input: head = [0] Output: 0
Example 3:
Input: head = [1] Output: 1
Example 4:
Input: head = [1,0,0,1,0,0,1,1,1,0,0,0,0,0,0] Output: 18880
Example 5:
Input: head = [0,0] Output: 0
Constraints:
- The Linked List is not empty.
- Number of nodes will not exceed
30
. - Each node's value is either
0
or1
.
Related Topics:
Linked List, Bit Manipulation
Solution 1.
// OJ: https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int getDecimalValue(ListNode* head) {
int ans = 0;
while (head) {
ans = (ans << 1) + head->val;
head = head->next;
}
return ans;
}
};
Java
-
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ class Solution { public int getDecimalValue(ListNode head) { int decimalValue = 0; ListNode temp = head; while (temp != null) { decimalValue <<= 1; decimalValue += temp.val; temp = temp.next; } return decimalValue; } }
-
// OJ: https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/ // Time: O(N) // Space: O(1) class Solution { public: int getDecimalValue(ListNode* head) { int ans = 0; for (; head; head = head->next) ans = ans << 1 | head->val; return ans; } };
-
# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def getDecimalValue(self, head: ListNode) -> int: ans = 0 while head: ans = ans << 1 | head.val head = head.next return ans