LintCode Subarray Sum

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.
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Example
Given [-3, 1, 2, -3, 4], return [0, 2] or [1, 3].
Note
There is at least one subarray that it's sum equals to zero.

卡了一会，需要求一个sum = sigma(a[0]..a[i])的累加和，因为如果某个subarray是0的话，就会出现有两个累加和相同的请求。比如上面例子中的累加和：

-3 -2 0 -3 -1

可以看到有两个-3，也就是说在这两者之间其subarray和为0，才导致了两个位置累加和重叠。还有一种简单的情况就是碰到摸个累加和为0，那么也就是说从数组开始到当前位置的和为0。

class Solution {
public:
    /**
     * @param nums: A list of integers
     * @return: A list of integers includes the index of the first number 
     *          and the index of the last number
     */
    vector<int> subarraySum(vector<int> nums){
        // write your code here
        
        int len = nums.size();
        vector<int> res;
 
        int sum = 0;
    
        unordered_map<int, int> val2idx;
        
        for (int i = 0; i < len; i++) {
            sum += nums[i];
            if (sum == 0) {
                res = {0, i};
                return res;
            }
            if (val2idx.count(sum) > 0) {
                res = {val2idx[sum] + 1, i};
                break;
            }
            val2idx[sum] = i;
        }
        
        return res;
    }
};

感觉放在easy等级里不合适