1. Two Sum

Easy

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,

从头开始，准备使用Java再刷一遍题，Java 0基础，syntax，数据结构，慢慢感觉。

1. Brute force

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] ans = new int[2];
        for(int i=0; i<nums.length; i++){
            for(int j=i+1; j<nums.length; j++){
                if(nums[i]+nums[j]==target){
                    ans[0] = i;
                    ans[1] = j;
                }
            }
        }
        return ans;
    }
}

Syntax:

1. java和c++很像，都需要提前声明变量类型

2. for循环的变量一定要记得声明类型

3. new https://blog.csdn.net/ljheee/article/details/52235915

   Point originOne = new Point(23, 94);

   Rectangle rectOne = new Rectangle(originOne, 100, 200);

   Rectangle rectTwo = new Rectangle(50, 100);

   第一行创建了一个 Point 类的对象，第二个和第三个线创建一个Rectangle 矩形类的对象。

   这些陈述中的每一个都有三个部分（详细讨论）：

   声明Declaration：粗体代码是将变量名称与对象类型关联的变量声明。

   实例化Instantiating ：new关键字是一个java运算符，它用来创建对象。

   初始化Initialization：new运算符，随后调用构造函数(interface)，初始化新创建的对象。

4. Hash map

class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer, Integer> mem = new HashMap();
        for(int i=0; i<nums.length; i++){
            int sub = target-nums[i];
            if(mem.containsKey(sub)){
                return new int[]{i, mem.get(sub)};
            }
            mem.put(nums[i], i);
        }
        throw new IllegalArgumentException("No two sum solution");
    }
}

class Solution {
    public int[] twoSum(int[] nums, int target) {
        HashMap<Integer, Integer> mem = new HashMap();
        for(int i=0; i<nums.length; i++){
            int sub = target-nums[i];
            if(mem.containsKey(sub)){
                return new int[]{i, mem.get(sub)};
            }
            mem.put(nums[i], i);
        }
        return new int[]{};
    }
}