import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
/**
* Source : https://oj.leetcode.com/problems/add-two-numbers/
*
* Created by lverpeng on 2017/6/23.
*
* You are given two linked lists representing two non-negative numbers.
* The digits are stored in reverse order and each of their nodes contain a single digit.
* Add the two numbers and return it as a linked list.
*
* Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
* Output: 7 -> 0 -> 8
*
*/
public class AddTwoNumbers {
/**
* 问题就是求两个数的和,只不过这两个数分别存在两个链表中,低位在前,高位在后,求出的和也存在链表中
* 循环较长的那个链表:
* 如果是当前有个链表都有值,则将两个数与上次计算的商(也就是进位)求和,结果的对于10的余数存于结果链表对一个的位置,结果对于10的商保存在变量中(也就是进位),
* 如果已经超过了其中一个链表的长度,则直接把另一个链表的数存入结果链表
*
* @param num1
* @param num2
* @return
*/
public List<Integer> addTwoNumbers (List<Integer> num1, List<Integer> num2) {
List<Integer> numbers = new ArrayList<Integer>();
int size = 0;
if (num1.size() > num2.size()) {
size = num1.size();
} else {
size = num2.size();
}
int carry = 0;
int sum = 0;
for (int i = 0; i < size; i++) {
if (num1.size() > i && num2.size() > i) {
sum = carry + num1.get(i) + num2.get(i);
} else if (num1.size() - 1 > i) {
sum = carry + num1.get(i);
} else {
sum = carry + num2.get(i);
}
numbers.add(sum % 10);
carry = sum / 10;
}
if (carry > 0) {
numbers.add(carry);
}
return numbers;
}
public List<Integer> addTwoNumbersRefactor (List<Integer> num1, List<Integer> num2) {
List<Integer> result = new ArrayList<Integer>();
int size = num1.size() > num2.size() ? num1.size() : num2.size();
int carry = 0;
int sum = 0;
for (int i = 0; i < size; i++) {
sum = carry + getNumber(num1, i) + getNumber(num2, i);
carry = sum / 10;
result.add(sum % 10);
}
if (carry > 0) {
result.add(carry);
}
return result;
}
private int getNumber(List<Integer> numbers, int i) {
if (i < 0 || i >= numbers.size()) {
return 0;
}
return numbers.get(i);
}
public static void main(String[] args) {
AddTwoNumbers addTwoNumbers = new AddTwoNumbers();
List<Integer> num1 = new ArrayList<Integer>(){{
add(2);
add(4);
add(3);
}};
List<Integer> num2 = new ArrayList<Integer>(){{
add(5);
add(6);
add(4);
}};
System.out.println(addTwoNumbers.addTwoNumbers(num1, num2));
System.out.println(addTwoNumbers.addTwoNumbersRefactor(num1, num2));
}
}