Example
Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

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

時間複雜度o(n),空間複雜度n
這裡陣列是沒有順序的，所以不能直接從利用從首和尾部同時遍歷的方法
不過可以加點條件來實現

 public int[] twoSum(int[] nums, int target) {
        int[] nums2 = nums.clone();//克隆
        Arrays.sort(nums2);
        int first = 0, second = nums.length - 1;
        while (nums2[first] + nums2[second] != target) {
            if (nums2[first] + nums2[second] > target)
                second--;
            else
 

                first++;
        }
        int k = 0;
        int[] result = {0, 0};
        for (int i = 0; i < nums.length; i++)
            if (nums[i] == nums2[first] || nums[i] == nums2[second])
                result[k++] = i;
        return result;
    }

在leetcode上第二個方法還快了1毫秒