[LintCode] Subarray Sum | 前缀和 + Hash Map

http://www.lintcode.com/zh-cn/problem/subarray-sum/#

思路1：Brute Force

最容易想到的方法就是暴力枚举子数组，计算和，并返回和为0的子数组。

Time complexity: O(n^2)
Space complexity: O(1)

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){
        vector<int> ret;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            int sum = 0;
            for (int j = i; j < n; ++j) {
                sum += nums[j];
                if (sum == 0) {
                    ret.push_back(i);
                    ret.push_back(j);
                    return ret;
                }
            }
        }
        return ret;
    }
};

思路2：Prefix Sum + Hash Map

扫描一次数组，跟踪前缀和并用一个Hash Map保存(前缀和, 结尾位置)的key-value pair。每到一个位置i，计算前缀和sum，并判断：

1. sum是否为0，如果是，记录区间并返回
2. 在Hash Map中是否有key为sum的key-value pair，如果有，设这样的key-value pair为(sum, j)，和为0的子数组为nums[j+1...i]记录区间并返回

Time complexity: O(n)
Space complexity: O(n)

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){
        vector<int> ret;
        unordered_map<long long, int> mymap;
        unordered_map<long long, int>::iterator it;
        long long sum = 0;
        for (int i = 0; i < nums.size(); ++i) {
            sum += nums[i];
            if (sum == 0) {
                ret.push_back(0);
                ret.push_back(i);
                return ret;
            }
            it = mymap.find(sum);
            if (it != mymap.end()) {
                ret.push_back(it->second + 1);
                ret.push_back(i);
                return ret;
            } else {
                mymap[sum] = i;
            }
        }
        return ret;
    }
};

上面的代码可以简化，在循环之前加上maymap[0] = -1就可以将判断1和2统一起来：

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){
        vector<int> ret;
        unordered_map<long long, int> mymap;
        unordered_map<long long, int>::iterator it;
        mymap[0] = -1;  // 加上这句可以简化代码
        long long sum = 0;
        for (int i = 0; i < nums.size(); ++i) {
            sum += nums[i];
            it = mymap.find(sum);
            if (it != mymap.end()) {
                ret.push_back(it->second + 1);
                ret.push_back(i);
                return ret;
            }
            mymap[sum] = i;
        }
        return ret;
    }
};