1.Two Sum

决心把leetcdoe从头做一遍， 就从 two sum 开始吧

本题的提意思非常直观，简单说一下
给了一个数组 nums 和 target, 找出数组的index 使其值的和等于target
Eg. nums = [2, 7, 11, 15], target = 9;
return [0, 1] (因为nums[0] + nums[1] = 9)

本题最直接的方法是用brute force来做，用两个for loop 从头到尾遍历一遍

    public int[] twoSumBruteForce(int[] nums, int target) {
        if (nums == null || nums.length < 2) {
            throw new IllegalArgumentException("argument is not valid");
        }
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    return new int[]{i, j};
                }
            }
        }
        throw new RuntimeException("index not found");
    }

我们也可以用 Hashmap 来做
遍历数组
key存当前的数组的值
value存当前数组的index
如果target - nums[i] 在map中存在key
说明我们找到了目标index

    public int[] twoSum(int[] nums, int target) {
        if (nums == null || nums.length < 2) {
            throw new IllegalArgumentException("argument is not valid");
        }
        HashMap map = new HashMap<>();

        for (int i = 0; i < nums.length; i++) {
            if (map.containsKey(target - nums[i])) {
                 return new int[] {map.get(target - nums[i]), i};
            }
            map.put(nums[i], i);
        }
        throw new RuntimeException("index not found");
    }