1. Two Sum

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].

一刷：任意两个数的和等于目标值，首先想到的是指定数组分成n*(n-1)个组合，判断这两个数的数值是否等于目标值

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) {
                    result[0] = i;
                    result[1] = j;
                    return result;
                }
            }
        }
        return result;
    }

二刷：可以考虑用map来获取第二个数是否存在，如果存在则打印出第二个数的索引及第一个数的索引

public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        int length = nums.length;
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < length; i++) {
            if (map.containsKey(target - nums[i])) {
                result[0] = map.get(target - nums[i]);
                result[1] = i;
                return result;
            } else {
                map.put(nums[i], i);
            }
        }
        return result;
    }