leetcode：two sum，Subarray Sum类的问题 （hash table)

560. Subarray Sum Equals K
     https://leetcode.com/problems/subarray-sum-equals-k/description/
     Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Tips：
1，init res = 0. or
Local variables aren’t initialized unless you do it yourself. What you’re seeing is whatever garbage was on the stack before your method was called.
2, init the summap[0] to 1 to cover the case that: sumsofar = k.
3, hash里面村的是前面sumsofar出现的次数，要寻找的是sumsofar - k，因为目标是：
后面的sumsofar - 前面的sumsofar = k，前面的sunsofar都存在hash里

class Solution {
public:
    int subarraySum(vector<int>& nums, int k) {
        unordered_map<int,int>summap;
        int res = 0;
        summap[0] = 1;
        int sumsofar = 0;
        for(auto num : nums)
        {
            sumsofar += num;
            res += summap[sumsofar - k];
            summap[sumsofar]++;            
        }
        return res;
    }
};

1. Two Sum
   https://leetcode.com/problems/two-sum/description/

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int>map;
        vector<int>res;
        for(int i= 0; i < nums.size(); i++)
        {        
            int candi = target - nums[i];
            if(map.find(candi) != map.end())
            {
                res.push_back(map[candi]);
                res.push_back(i);
                break;
            }
            map[nums[i]] = i;
        }
        
        return res;
    }
};

523. Continuous Subarray Sum
     https://leetcode.com/problems/continuous-subarray-sum/description/
     Given a list of non-negative numbers and a target integer k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of k, that is, sums up to n*k where n is also an integer.

这个题更有意思，加了限制条件，
1，at least 2 subarray, 存入一个pre，保证现在的sumsofar和前一个sunsofar中间有个gap。现在的sumsofar这个cycle里要先放到pre里，等下一个cycle才能放到hash里，然后再下一个cycle才能找到。
cornor case: [0,1,0] 0
2，目标是k的整数倍：存入set的是sumsofar % kd结果，如果同样的mod结果已经出现过的话，说明：现在的sumsofar-前面已经有sumsofar一定是k的倍数。
3，只需要输出true or false，所以不需要unordered_map, unordered_set足够了

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        unordered_set<int>summap;
        int sumsofar = 0;
        int pre = 0;
        
        for(auto num : nums)
        {
            sumsofar += num;
            sumsofar = k == 0? sumsofar : sumsofar % k;
            if(summap.count(sumsofar))
            {
                return true;
            }
            summap.insert(pre);
            pre = sumsofar;            
        }
        
        return false;
    }
};