2.5 You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list. EXAMPLE Input: (3 -> 1 -> 5) + (5 -> 9 -> 2) Output: 8 -> 0 -> 8
给两个整型链表，每个节点包含一个位数，这些位数是反向存放的，也就是个位数在链表首部，编写程序，对这两个整数求和，并用链表形式返回结果。
进阶：
假设这些位数是正向存放的，请再做一遍。
1.自己的思路是：取每个list的第一个，相加（随时取余，相除），放入新建一个列表的最后一个。
1 public class newAdd5 { 2 public static LinkedList<Integer> sum1(LinkedList<Integer> l1, 3 LinkedList<Integer> l2) { 4 LinkedList<Integer> l3 = new LinkedList<Integer>(); 5 if (l1.isEmpty()) 6 return l2; 7 if (l2.isEmpty()) 8 return l1; 9 int carry = 0; 10 while (!l1.isEmpty() && !l2.isEmpty()) { 11 int sum = 0; 12 sum = (l1.peek() + l2.peek() + carry) % 10; 13 carry = (l1.poll() + l2.poll() + carry) / 10; 14 l3.addFirst(sum); 15 } 16 while (!l1.isEmpty() && l2.isEmpty()) { 17 int sum = 0; 18 sum = (l1.peek() + carry) % 10; 19 carry = (l1.poll() + carry) / 10; 20 l3.addFirst(sum); 21 } 22 while (!l2.isEmpty() && l1.isEmpty()) { 23 int sum = 0; 24 sum = (l2.peek() + carry) % 10; 25 carry = (l2.poll() + carry) / 10; 26 l3.addFirst(sum); 27 } 28 return l3; 29 30 } 31 32 // follow up 33 public static LinkedList<Integer> sum2(LinkedList<Integer> l1, 34 LinkedList<Integer> l2) { 35 java.util.Collections.reverse(l1); 36 java.util.Collections.reverse(l2); 37 LinkedList<Integer> l3 = new LinkedList<Integer>(); 38 if (l1.isEmpty()) 39 return l2; 40 if (l2.isEmpty()) 41 return l1; 42 int carry = 0; 43 while (!l1.isEmpty() && !l2.isEmpty()) { 44 int sum = 0; 45 sum = (l1.peek() + l2.peek() + carry) % 10; 46 carry = (l1.poll() + l2.poll() + carry) / 10; 47 l3.addFirst(sum); 48 } 49 while (!l1.isEmpty() && l2.isEmpty()) { 50 int sum = 0; 51 sum = (l1.peek() + carry) % 10; 52 carry = (l1.poll() + carry) / 10; 53 l3.addFirst(sum); 54 } 55 while (!l2.isEmpty() && l1.isEmpty()) { 56 int sum = 0; 57 sum = (l2.peek() + carry) % 10; 58 carry = (l2.poll() + carry) / 10; 59 l3.addFirst(sum); 60 } 61 return l3; 62 63 } 64 65 public static void main(String[] args) { 66 Random rd = new Random(); 67 LinkedList<Integer> list1 = new LinkedList<Integer>(); 68 LinkedList<Integer> list2 = new LinkedList<Integer>(); 69 int n1 = rd.nextInt(5); 70 for (int i = 0; i < n1; i++) { 71 list1.add(rd.nextInt(10)); 72 } 73 int n2 = rd.nextInt(6); 74 for (int i = 0; i < n2; i++) { 75 list2.add(rd.nextInt(10)); 76 } 77 System.out.println(list1.toString()); 78 System.out.println(list2.toString()); 79 System.out.println("After Adding: "); 80 System.out.println(sum1(list1, list2).toString()); 81 LinkedList<Integer> list3 = new LinkedList<Integer>(); 82 LinkedList<Integer> list4 = new LinkedList<Integer>(); 83 for (int i = 0; i < n1; i++) { 84 list3.add(rd.nextInt(10)); 85 } 86 for (int i = 0; i < n2; i++) { 87 list4.add(rd.nextInt(10)); 88 } 89 System.out.println("Follow Up: "); 90 System.out.println(list3.toString()); 91 System.out.println(list4.toString()); 92 System.out.println(sum2(list3, list4).toString()); 93 } 94 }
然后在网上看的方法更简洁：
public class Sum5 { public static void main(String[] args) { Random rd = new Random(); LinkedList<Integer> list1 = new LinkedList<Integer>(); LinkedList<Integer> list2 = new LinkedList<Integer>(); int n1 = rd.nextInt(5); for (int i = 0; i < n1; i++) { list1.add(rd.nextInt(10)); } int n2 = rd.nextInt(6); for (int i = 0; i < n2; i++) { list2.add(rd.nextInt(10)); } System.out.println(list1.toString()); System.out.println(list2.toString()); System.out.println("After Adding: "); System.out.println(addReverse(list1, list2).toString()); LinkedList<Integer> list3 = new LinkedList<Integer>(); LinkedList<Integer> list4 = new LinkedList<Integer>(); for (int i = 0; i < n1; i++) { list3.add(rd.nextInt(10)); } for (int i = 0; i < n2; i++) { list4.add(rd.nextInt(10)); } System.out.println("Follow Up: "); System.out.println(list3.toString()); System.out.println(list4.toString()); System.out.println(addForward(list3, list4).toString()); } private static LinkedList<Integer> addReverse(LinkedList<Integer> list1, LinkedList<Integer> list2) { LinkedList<Integer> list3 = new LinkedList<Integer>(); int sum = 0; while (!list1.isEmpty() || !list2.isEmpty() || sum != 0) { int tempsum = sum; if (!list1.isEmpty()) { tempsum += list1.poll(); } if (!list2.isEmpty()) { tempsum += list2.poll(); } list3.addFirst(tempsum % 10); sum = tempsum / 10; } return list3; } private static LinkedList<Integer> addForward(LinkedList<Integer> list1, LinkedList<Integer> list2) { LinkedList<Integer> list3 = new LinkedList<Integer>(); int sum = 0; java.util.Collections.reverse(list1); java.util.Collections.reverse(list2); while (!list1.isEmpty() || !list2.isEmpty() || sum != 0) { int tempsum = sum; if (!list1.isEmpty()) { tempsum += list1.poll(); } if (!list2.isEmpty()) { tempsum += list2.poll(); } list3.addFirst(tempsum % 10); sum = tempsum / 10; } return list3; } }