剑指 Offer II 025. 链表中的两数相加（445. 两数相加 II【进阶版本】 && 2. 两数相加）

题目：

 

 

 

思路：

【1】剑指 Offer II 025. 链表中的两数相加（445. 两数相加 II【进阶版本】）

【1.1】链表相加，反转链表之后再相加其实是最为靠谱的

【1.2】基于进阶说不给反转，那么要么采用递归，从底层往上走。（要么就是全部取出值来形成字符串相加）

【2】2. 两数相加

代码展示：

【1】剑指 Offer II 025. 链表中的两数相加（445. 两数相加 II【进阶版本】）

基于反转链表的方式：

//时间1 ms击败100%
//内存41.3 MB击败85.6%
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode tem1 = l1, tem2 = l2;
        int index1 = 0, index2 = 0;
        while (tem1 != null || tem2 != null){
            if (tem1 != null) {
                tem1 = tem1.next;
                index1++;
            }
            if (tem2 != null){
                tem2 = tem2.next;
                index2++;
            }
        }
        ListNode res = addTwoIndexNumbers(l1,index1,l2,index2);
        if (res.val >= 10) {
            ListNode ans = new ListNode(res.val / 10);
            res.val = res.val % 10;
            ans.next = res;
            res = ans;
        }
        return res;
    }

    public ListNode addTwoIndexNumbers(ListNode l1, int index1, ListNode l2, int index2) {
        if (index1 == 1 && index2 == 1) return new ListNode(l1.val + l2.val);
        int sum = 0 , val1 = 0, val2 = 0;
        ListNode next, res;
        if (index1 > index2) {
            next = addTwoIndexNumbers(l1.next,index1 - 1,l2,index2);
            val1 = l1.val;
        }else if (index1 < index2){
            next = addTwoIndexNumbers(l1, index1,l2.next,index2 - 1);
            val2 = l2.val;
        }else{
            next = addTwoIndexNumbers(l1.next,index1 - 1,l2.next,index2 - 1);
            val1 = l1.val;
            val2 = l2.val;
        }
        if (next.val >= 10){
            sum = next.val / 10;
            next.val = next.val % 10;
        }
        res = new ListNode(val1 + val2 + sum);
        res.next = next;
        return res;
    }
}

基于字符串相加的方式：

//时间2 ms击败68.99%
//内存41.6 MB击败39.48%
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode tem1 = l1, tem2 = l2;
        StringBuffer buf1 = new StringBuffer();
        StringBuffer buf2 = new StringBuffer();
        while (tem1 != null || tem2 != null){
            if (tem1 != null) {
                buf1.append(tem1.val);
                tem1 = tem1.next;
            }
            if (tem2 != null){
                buf2.append(tem2.val);
                tem2 = tem2.next;
            }
        }
        int size1 = buf1.length(), size2 = buf2.length();
        ListNode res = null;
        int diff = 0;
        while (size1 > 0 || size2 > 0){
            int val1 = 0 , val2 = 0;
            if (size1 > 0) val1 = buf1.charAt(--size1) - '0';
            if (size2 > 0) val2 = buf2.charAt(--size2) - '0';
            int sum = val1 + val2 + diff;
            //差值用了就要清除
            diff = 0;
            if (sum > 9) {
                diff = sum / 10;
                sum = sum % 10;
            }
            ListNode head = new ListNode(sum);
            head.next = res;
            res = head;
        }
        //如果最后还存在，证明是要增加一个头部节点的
        if (diff != 0) {
            ListNode ans = new ListNode(diff);
            ans.next = res;
            res = ans;
        }
        return res;
    }

}

 

基于递归的方式相加：

//时间1 ms击败100%
//内存41.1 MB击败97.19%
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) return l2;
        if (l2 == null) return l1;
        ListNode tem1 = l1, tem2 = l2;
        int index1 = 0, index2 = 0;
        while (tem1 != null || tem2 != null){
            if (tem1 != null) {
                tem1 = tem1.next;
                index1++;
            }
            if (tem2 != null){
                tem2 = tem2.next;
                index2++;
            }
        }
        ListNode res = addTwoIndexNumbers(l1,index1,l2,index2);
        if (res.val >= 10) {
            ListNode ans = new ListNode(res.val / 10);
            res.val = res.val % 10;
            ans.next = res;
            res = ans;
        }
        return res;
    }

    public ListNode addTwoIndexNumbers(ListNode l1, int index1, ListNode l2, int index2) {
        if (index1 == 1 && index2 == 1) return new ListNode(l1.val + l2.val);
        int sum = 0 , val1 = 0, val2 = 0;
        ListNode next, res;
        if (index1 > index2) {
            next = addTwoIndexNumbers(l1.next,index1 - 1,l2,index2);
            val1 = l1.val;
        }else if (index1 < index2){
            next = addTwoIndexNumbers(l1, index1,l2.next,index2 - 1);
            val2 = l2.val;
        }else{
            next = addTwoIndexNumbers(l1.next,index1 - 1,l2.next,index2 - 1);
            val1 = l1.val;
            val2 = l2.val;
        }
        if (next.val >= 10){
            sum = next.val / 10;
            next.val = next.val % 10;
        }
        res = new ListNode(val1 + val2 + sum);
        res.next = next;
        return res;
    }
}

 【2】2. 两数相加（这种由于是逆序的，直接遍历相加然后进位即可）

//时间1 ms 击败 100%
//内存42.2 MB 击败 14.80%
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        //预先设置0节点
        ListNode n1 = l1 , n2 = l2 ,zeroNode = new ListNode(0);
        ListNode tem , resNode = new ListNode(-1);
        ListNode head = resNode;
        int carryBit = 0;
        // 因为是逆序存储的所以可以直接加
        while (n1 != null || n2 != null){
            if (n1 == null) n1 = zeroNode;
            if (n2 == null) n2 = zeroNode;
            int value = n1.val+n2.val+carryBit;
            carryBit = value > 9 ? value / 10 : 0 ;
            value = value - carryBit*10;
            tem = new ListNode(value);
            head.next = tem;
            head = head.next;
            // 链表移动
            n1 = n1.next;
            n2 = n2.next;
        }
        if (carryBit > 0){
            tem = new ListNode(carryBit);
            head.next = tem;
        }
        return resNode.next;
    }
}