【一次过】Lintcode 786. 链表的反向带权和

给定一个链表，求出这个链表的带权和。链表结点的权值指的是该结点到链表尾节点的结点数。

样例

样例1

输入: 3 -> 2 -> 5 -> 1
输出: 29
解释:
(3 * 4 + 2 * 3 + 5 * 2 + 1) = 29

样例2

输入: 1 -> 2 -> 3 -> 4
输出: 20
解释:
(1 * 4 + 2 * 3 + 3 * 2 + 4) = 20

---

解题思路：

先求链表长度，然后遍历链表。

/**
 * Definition for ListNode
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param head: the given linked list
     * @return:  the weighted sum in reverse order
     */
    public int weightedSumReverse(ListNode head) {
        int res = 0;
        
        int lens = listLength(head);
        
        while(lens != 0){
            res += head.val * lens;
            
            lens--;
            head = head.next;
        }
        
        return res;
    }
    
    private int listLength(ListNode head){
        int lens = 0;
        
        while(head != null){
            lens++;
            head = head.next;
        }
        
        return lens;
    }
}