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

Constraints:

• The Linked List is not empty.
• Number of nodes will not exceed 30.
• Each node's value is either 0 or 1.

这道题让把一个用链表表示的二进制数转为一个整型数，链表的结点值只有0或1，而且首结点是二进制数的最高位。这题主要考察两点，一个是二进制数如何转十进制数，另一个是遍历链表。都不是太难，直接遍历链表，每次先将 res 自乘以2，因为新加一个结点，说明之前的每一位都要左移一位，所以要乘以2，然后再加上当前结点值，同时将 head 指针右移一位即可，参见代码如下：

class Solution {
public:
    int getDecimalValue(ListNode* head) {
        int res = 0;
        while (head) {
            res = res * 2 + head->val;
            head = head->next;
        }
        return res;
    }
};

Github 同步地址:

#1290

参考资料：

https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/

https://leetcode.com/problems/convert-binary-number-in-a-linked-list-to-integer/discuss/629087/Detailed-explanation-Java-%3A-faster-than-100.00

LeetCode All in One 题目讲解汇总(持续更新中...)

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