算法---从一个数组(或者集合中)找出和为某个值的下标

需求，例如：

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

    public static void main(String[] args) {        int[] nums = new int[]{2, 7, 5, 1, 3};        int[] ints = twoSum(nums, 10);        System.out.println(ints[0]);        System.out.println(ints[1]);    }

    public static int[] twoSum(int[] nums, int target) {        //因为你要找到这两个相加等于目标数,因此我认为你至少要遍历一次        Map hashMap = new HashMap<>();        for (int i = 0; i < nums.length; i++) {            int complement = target - nums[i];            if (hashMap.containsKey(complement)) {                return new int[]{hashMap.get(complement), i};            }            hashMap.put(nums[i], i);        }        throw new IllegalArgumentException("No two sum solution");    }

public static int[] twoSum(int[] nums, int target) {        for (int i = 0; i < nums.length; i++) {            for (int j = i + 1; j < nums.length; j++) {                if (nums[j] == target - nums[i]) {                    return new int[]{i, j};                }            }        }        throw new IllegalArgumentException("No two sum solution");    }