#138 Subarray Sum

题目描述：

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.

Have you met this question in a real interview? 

Yes

Example

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

题目思路：

首先建立一个presum array，对于每个presum中的元素i，它的value为sum of all elements <= i in nums array.那么要求subarray where the sum of numbers is zero,那就意味着求两个index i和j，使presum[i] = presum[j]。在不断算presum的过程中，用一个map把之前算过的presum value和它对应的index存起来。当当前的presum在map中能找到时，就找到了两个符合题意的index。

Mycode（AC = 1842ms）：

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        vector<int> ans(2, 0);        if (nums.size() <= 1) return ans;                vector<int> presum(nums.size() + 1, 0);        map<int, int> helper;        helper[presum[0]] = 0;                 // use a map to store the presum value and        // its index which appeared before        for (int i = 1; i <= nums.size(); i++) {            presum[i] = presum[i - 1] + nums[i - 1];            if (helper.find(presum[i]) != helper.end()) {                ans[0] = helper[presum[i]];                ans[1] = i - 1;                break;            }            else {                helper[presum[i]] = i;            }        }                return ans;    }};