1. 36进制大整数相加

两个36进制的大整数以字符串的形式给出，求出两个大整数的和，并以字符串方式输出。（头条面试题）

比如：12346 + GSFTYHS = GSGW1LY

public class Format36 {    public static void main(String[] args) {        System.out.println(sum("12346", "GSFTYHS"));    }        public static String sum(String num1, String num2){        String ascii = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";        if(num1.length() > num2.length()){            StringBuilder sb = new StringBuilder(num2);            for (int i = 0; i < num1.length() - num2.length(); i++) {                sb.insert(0, '0');            }            num2 = sb.toString();        }        if(num1.length() < num2.length()){            StringBuilder sb = new StringBuilder(num1);            for (int i = 0; i < num2.length() - num1.length(); i++) {                sb.insert(0, '0');            }            num1 = sb.toString();        }                int add = 0;//进位        StringBuilder sb = new StringBuilder();        for (int i = num1.length() - 1; i >= 0; i--) {            int n1 = ascii.indexOf(num1.charAt(i));            int n2 = ascii.indexOf(num2.charAt(i));            int sum = n1 + n2 + add;            System.out.println(sum);            sb.insert(0, ascii.charAt(sum % 36));            add = sum / 36;        }        if(add != 0)            sb.insert(0, add);        return sb.toString();    }}

2. 两个单链表求和

镜像：

给定两个非空链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储单个数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

进阶:

如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。

示例:

输入: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)输出: 7 -> 8 -> 0 -> 7

LeetCode：

思路：

用栈实现：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        Stack
     
       stack1 = new Stack<>();        Stack
      
        stack2 = new Stack<>();        while (l1 != null) {            stack1.push(l1);            l1 = l1.next;        }        while (l2 != null) {            stack2.push(l2);            l2 = l2.next;        }        int add = 0;// 进位        int value1 = 0;        int value2 = 0;        ListNode ago = null;        ListNode node = null;        while (!stack1.isEmpty() || !stack2.isEmpty()) {            if (!stack1.isEmpty()) {                value1 = stack1.pop().val;            } else {                value1 = 0;            }            if (!stack2.isEmpty()) {                value2 = stack2.pop().val;            } else {                value2 = 0;            }            int sum = value1 + value2 + add;            node = new ListNode(sum % 10);            add = sum / 10;            node.next = ago;            ago = node;        }        if (add > 0) {            node = new ListNode(add);            node.next = ago;            ago = node;        }        return node;    }
      
     

反转单链表实现：

public ListNode addTwoNumbers2(ListNode l1, ListNode l2) {        l1 = reverseList(l1);        l2 = reverseList(l2);        int add = 0;// 进位        int value1 = 0;        int value2 = 0;        ListNode ago = null;        ListNode node = null;        while (l1 != null || l2 != null) {            if (l1 != null) {                value1 = l1.val;                l1 = l1.next;            } else {                value1 = 0;            }            if (l2 != null) {                value2 = l2.val;                l2 = l2.next;            } else {                value2 = 0;            }            int sum = value1 + value2 + add;            node = new ListNode(sum % 10);            add = sum / 10;            node.next = ago;            ago = node;        }        if (add > 0) {            node = new ListNode(add);            node.next = ago;            ago = node;        }                //把链表改回来        l1 = reverseList(l1);        l2 = reverseList(l2);        return node;    }        public ListNode reverseList(ListNode head){        ListNode pre = null;        ListNode next = null;        while(head != null){            next = head.next;            head.next = pre;            pre = head;            head = next;        }        return pre;    }