突发奇想，感觉自己应该提升一下算法的能力，怎么办呢，来刷LeetCode吧！
Leet Code OJ 1. Two Sum
题目：
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.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
翻译：
给定一个整形数组和一个整数target，返回2个元素的下标，它们满足相加的和为target。
你可以假定每个输入，都会恰好有一个满足条件的返回结果。
Java版代码1（时间复杂度O(n)）：
public class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer,Integer> map=new HashMap<>();
for(int i=0;i<nums.length;i++){
Integer index=map.get(target-nums[i]);
if(index==null){
map.put(nums[i],i);
}else{
return new int[]{i,index};
}
}
return new int[]{0,0};
}
}
Java版代码2（时间复杂度O(n^2)）：
public class Solution {
public int[] twoSum(int[] nums, int target) {
int[] result=new int[2];
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 result;
}
}
Leet Code OJ 2. Add Two Numbers
题目：
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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
给出两个表示两个非负整数的非空链表。数字以相反的顺序存储，它们的每个节点都包含一个数字。加上这两个数并返回一个链表。
你可以假设这两个数字不包含任何前导零，除了第0个数字本身。
