leetcode 1290. Convert Binary Number in a Linked List to Integer（链表的二进制转为十进制数）

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.

Input: head = [1,0,1]
Output: 5
Explanation: (101) in base 2 = (5) in base 10

把链表对应的二进制数转为十进制数，链表的head对应二进制数的最高位。

思路：
可能会想到用一个数组把数字都保存下来，然后从低位开始转换。但是这样space 和 time complexity都很高。

可以用二进制位运算解决，定义一个变量result。
从head读起，每往右读一位，result左移一位，同时与当前head.val取或，这相当于result*2 + head.val，直到head == null

    public int getDecimalValue(ListNode head) {
        if(head == null) {
            return 0;
        }
        
        int result = 0;
        
        while(head != null) {
            result = result << 1 | head.val;
            head = head.next;
        }
        
        return result;
    }