因为set的无序 所以在获取第一个数的下标的时候会有问题，所以这里使用set不是很好，不过可以使用map，用来存储下标和数：

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

        return ans;
    }

难道不是用map更好？
go版本实现：
func twoSum(arr []int, target int) [2]int{
    tmpMap := map[int]int{}
    for k, v := range arr{
        if index, ok := tmpMap[target-v]; ok{
            return [2]int{index, k}
        }
        tmpMap[v] = k
    }
    return [2]int{-1,-1}
}

此题HashSet和HashMap都可以，代码如下：
public int[] twoSum(int[] nums, int target) {

        if (nums == null) {
            return new int[]{};
        }

        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            Integer existIndex = map.get(nums[i]);
            if (existIndex != null) {
                if (target == 2 * nums[i]) {
                    return new int[]{existIndex, i};
                }
            } else {
                map.put(nums[i], i);
            }
        }

        for (Map.Entry<Integer, Integer> keyValue : map.entrySet()) {
            Integer key = map.get(target - keyValue.getKey());
            if (key != null && !key.equals(keyValue.getValue())) {
                return new int[]{keyValue.getValue(), key};
            }
        }
        return new int[]{};

    }


if (nums == null) {
            return new int[]{};
        }

        int[] ret = new int[2];
        HashSet<Integer> set = new HashSet<>();
        for (int i = 0; i < nums.length; i++) {
            int value = nums[i];
            if (set.contains(value) && target == value << 1) {
                ret[1] = i;
                for (int j = 0; j < i; j++) {
                    if (nums[j] == value) {
                        ret[0] = j;
                    }
                }
                return ret;
            }
            set.add(value);
        }

        int index = 0;
        for (int i = 0; i < nums.length; i++) {
            int value = nums[i];
            int targetValue = target - value;
            if (set.contains(targetValue) && value != targetValue) {
                ret[index++] = i;
            }
        }
        return ret;
    }

用Set不合适，改用map的话，key用来记录nums[i]的值，value用来记录对应的下标

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

private static int[] two_sum_method(int[] nums, int target) {
        Map<Integer, Integer> tempMap = new HashMap<Integer, Integer>();
        for (Integer i = 0; i < nums.length; i++) {
            if (tempMap.get(nums[i]) == null) {
                tempMap.put(target-nums[i], i);
            } else {
                return new int[]{i, tempMap.get(i)};
            }
        }
        return null;
    }

PHP 暴力解法

```
public function twoSum($nums, $target)
    {
        $outArr = [];
        foreach ($nums as $key => $num) {
            $num1 = $target - $num;
            foreach ($nums as $k => $n) {
                if ($k != $key) {
                    if ($num1 == $n) {
                        $outArr = [
                            $key,
                            $k,
                        ];
                        if (!empty($outArr)) {
                            return $outArr;
                        }
                    }
                }

            }
        }
        return $outArr;
    }
```

这段代码用了快排，时间复杂度n*logn，但是在leetcode上运行比时间复杂度N的C++快，亲测过，原因是什么？
public int[] twoSum(int[] nums, int target) {
         if(nums == null)
             return null;
         int[] nums2 = Arrays.copyOf(nums, nums.length);
         Arrays.sort(nums2);
         int a = 0, b = 0;
         int start = 0, end = nums2.length-1;
         //find two nums
         while(start<end){
             int sum = nums2[start] + nums2[end];
             if(sum < target)
                 start++;
             else if(sum > target)
                 end--;
             else{
                 a = nums2[start]; b = nums2[end];
                 break;
             }
         }
         //find the index of two numbers
         int[] res = new int[2];
         for(int i = 0; i < nums.length; i++){
             if(nums[i] == a){
                 res[0] = i;
                 break;
             }
         }
         if(a != b){
             for(int i = 0; i < nums.length; i++){
                 if(nums[i] == b){
                     res[1] = i;
                     break;
                 }
             }
         } else{
             for(int i = 0; i < nums.length; i++){
                 if(nums[i] == b && i != res[0]){
                     res[1] = i;
                     break;
                 }
             }
         }
         
         return res;
     }