Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].源码：
class Solution {
public int[] twoSum(int[] nums, int target) {
for (int i = 0; i < nums.length-1; i++){
for (int j = i+1; j < nums.length; j++){
if (nums[i] + nums[j] == target){
return new int[]{i, j};
}
}
}
return null;
}
}You are given two non-empty linked lists representing two non-negative integers. 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. You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.源码：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode headNode = null;
ListNode tailNode = null;
int numAdd = l1.val + l2.val;
headNode = new ListNode(numAdd%10);
ListNode l_temp1 = l1.next;
ListNode l_temp2 = l2.next;
tailNode = headNode;
while(l_temp1 != null || l_temp2 != null){
int num1;
int num2;
if (l_temp1 == null){
num1 = 0;
}else{
num1 = l_temp1.val;
l_temp1 = l_temp1.next;
}
if (l_temp2 == null){
num2 = 0;
}else{
num2 = l_temp2.val;
l_temp2 = l_temp2.next;
}
if (numAdd >= 10){
numAdd = num1 + num2 + 1;
}else {
numAdd = num1 + num2;
}
tailNode.next = new ListNode(numAdd%10);
tailNode = tailNode.next;
}
if (numAdd >= 10){
tailNode.next = new ListNode(1);
}
return headNode;
}
}