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# 138.Subarray Sum

## 1.Description(Easy)

Given an integer array, find a subarray where the sum of numbers is**zero**. Your code should return the index of the first number and the index of the last number.

### Notice

There is at least one subarray that it's sum equals to zero.

**Example**

Given`[-3, 1, 2, -3, 4]`, return`[0, 2]`or`[1, 3]`.

## 2.Code

记录每一个位置的sum，存入HashMap中，如果某一个sum已经出现过，那么说明中间的subarray的sum为0. 时间复杂度O(n)，空间复杂度O(n)

```
 public ArrayList<Integer> subarraySum(int[] nums) {
        ArrayList<Integer> result=new ArrayList<Integer>();
        if(nums==null ||nums.length==0){
            return result;
        }
        //key:sum  value:index
        HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();
        map.put(0, -1);

        int sum=0;
        for(int i=0;i<nums.length;i++){
            sum=sum+nums[i];
            if(map.containsKey(sum)){
                int start=map.get(sum)+1;
                int end=i;
                result.add(start);
                result.add(end);
                return result;
            }else{
                map.put(sum,i);
            }
        }
        return result;
    }
```