leetcode 523. Continuous Subarray Sum 560. Subarray Sum Equals K

下面两个题都是使用了unordered_map<> 来处理

数组的 连续子数组和 满足什么条件的问题。

523. Continuous Subarray Sum

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.

Example 1:

Input: [23, 2, 4, 6, 7],  k=6Output: TrueExplanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6Output: TrueExplanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.

Note:

1. The length of the array won't exceed 10,000.
2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.

最开始写的。TLE。

class Solution {public:    bool checkSubarraySum(vector<int>& nums, int k)     {        if (k < 0) k = -k;        int sum = 0;        unordered_map<int, int> ump;        ump[0] = -1;        for (int i = 0; i < nums.size(); i++)        {            sum += nums[i];            if (k == 0)            {                if (ump.find(sum) != ump.end() && i - ump[sum] >= 2)                     return true;            }            else            {                int n = 0;                while (n * k <= sum)                {                    if ( ump.find(sum - n * k) != ump.end() && i - ump[sum - n * k] >= 2)                         return true;                    n++;                }            }                        if (ump.find(sum) == ump.end())                ump[sum] = i;        }        return false;    }};

网上看了个方法。

sum_now % k = sum_old % k  《===》 sum_old - sum_now = n * k 关键！

class Solution {  public:      bool checkSubarraySum(vector<int>& nums, int k)     {          unordered_map<int, int> hash;          int sum = 0;           hash[0] = -1;          for (int i = 0; i < nums.size(); ++i)         {              sum += nums[i];              if(k) sum %= k; //如果k = 0,就一直不管余数的事            if (hash.find(sum) != hash.end())  //如果之前出现过这个余数,那就说明 sum_now % k = sum_old % k -> sum_old - sum_now =n*k            {                  if ( i - hash[sum] > 1) return true;              }              else                 hash[sum] = i;          }            return false;      }  }; 

560. Subarray Sum Equals K

Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k.

Example 1:

Input:nums = [1,1,1], k = 2Output: 2

Note:

1. The length of the array is in range [1, 20,000].
2. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].

看了一下数据范围，直接暴力解法就能A

class Solution {public:    int subarraySum(vector<int>& nums, int k)     {        int ret = 0;        for (int i = 0; i < nums.size(); i++)        {            int sum = 0;            for (int j = i; j < nums.size(); j++)            {                sum += nums[j];                if (sum == k) ret++;            }        }        return ret;       }};

另一种方法：这个方式非常巧妙，一定要记住！

用sum表示从数组开始位置到当前位置的数字相加和，用Hash Table来存储sum出现的次数，如果当前位置之前有相加和为sum-k的位置，则这两个位置之间的数字相加和为k，以当前位置结尾的相加和为k的子数组个数为hash[sum-k]，这样遍历整个数组即可得出满足条件的子数组个数。

class Solution {public:    int subarraySum(vector<int>& nums, int k)     {        int sum = 0, cnt = 0;          unordered_map<int, int> hash;            hash[0] = 1;          for(auto n:nums)         {              sum += n;              cnt += hash[sum-k]; // sum_now - sum_old = k 满足条件 -> sum_now - k = sum_old            hash[sum]++; //前面出现过的sum 需要加入map         }            return cnt;      }};