解法1：普通解法

    /**
     * @param numbers int整型一维数组
     * @param target  int整型
     * @return int整型一维数组
     */
    public int[] twoSum(int[] numbers, int target) {
        int[] res = new int[2];
        int pre, last;
        for (int i = 0; i < numbers.length; i++) {
            pre = numbers[i];
            last = target - pre;
            for (int j = i+1; j < numbers.length; j++) {
                if (last == numbers[j]) {
                    res[0] = i+1;
                    res[1] = j+1;
                }
            }
        }
        return res;
    }

解法2：高级解法

思路，尽量减少遍历。什么可以减少遍历？hash表。所以hashMap
hashMap也得遍历两次，一次把所有的值全扔进去，第二次才能开始找，那么有没有办法一次搞定？有，如下

    public int[] twoSum2(int[] numbers, int target) {
        Map<Integer, Integer> map = new HashMap<>();
        int n = numbers.length;
        for (int i = 0; i < n; i++) {
            int cur = numbers[i];
            int diff = target - cur;
            if (map.containsKey(diff)) {
                return new int[]{map.get(cur) + 1, i + 1};
            }
            map.put(cur, i);
        }
        throw new RuntimeException("result is not exist");
    }

这样写，可以，但是不够信达雅。改良一下，如下：

    public int[] twoSum2(int[] numbers, int target) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0, tmp; i < numbers.length; i++) {
            tmp = target - numbers[i];
            if (map.containsKey(tmp)) {
                return new int[]{map.get(tmp) + 1, i + 1};
            }
            map.put(numbers[i], i);
        }
        throw new RuntimeException("result is not exist");
    }