{"id":3127,"date":"2018-07-14T07:59:05","date_gmt":"2018-07-14T14:59:05","guid":{"rendered":"http:\/\/zxi.mytechroad.com\/blog\/?p=3127"},"modified":"2018-07-14T09:43:11","modified_gmt":"2018-07-14T16:43:11","slug":"leetcode-523-continuous-subarray-sum","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/hashtable\/leetcode-523-continuous-subarray-sum\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 523. Continuous Subarray Sum"},"content":{"rendered":"

Problem<\/strong><\/h1>\n

Given a list of\u00a0non-negative<\/b>\u00a0numbers and a target\u00a0integer<\/b>\u00a0k, write a function to check if the array has a continuous subarray of size at least 2 that sums up to the multiple of\u00a0k<\/b>, that is, sums up to n*k where n is also an\u00a0integer<\/b>.<\/p>\n

Example 1:<\/b><\/p>\n

Input:<\/b> [23, 2, 4, 6, 7],  k=6\r\nOutput:<\/b> True\r\nExplanation:<\/b> Because [2, 4] is a continuous subarray of size 2 and sums up to 6.\r\n<\/pre>\n
Example 2:<\/b><\/p>\n
Input:<\/b> [23, 2, 6, 4, 7],  k=6\r\nOutput:<\/b> True\r\nExplanation:<\/b> Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.\r\n<\/pre>\n
Note:<\/b><\/p>\n
    \n
  1. The length of the array won’t exceed 10,000.<\/li>\n
  2. You may assume the sum of all the numbers is in the range of a signed 32-bit integer.<\/li>\n<\/ol>\n
    Special case: <\/strong><\/h1>\n
    nums = [0,0], k = 0, return = True<\/p>\n
    Solution: Prefix Sum Reminder<\/strong><\/h1>\n
    Time complexity: O(n)<\/p>\n
    Space complexity: O(min(n, k))<\/p>\n
    \/\/ Author: Huahua\r\n\/\/ Running time: 16 ms (<99.62%)\r\nclass Solution {\r\npublic:\r\n  bool checkSubarraySum(vector<int>& nums, int k) {    \r\n    unordered_map<int, int> m; \/\/ sum % k -> first index\r\n    m[0] = -1;\r\n    int sum = 0;\r\n    for (int i = 0; i < nums.size(); ++i) {\r\n      sum += nums[i];\r\n      if (k != 0) sum %= k;      \r\n      if (m.count(sum)) {\r\n        if (i - m.at(sum) > 1) return true;\r\n      } else {\r\n        m[sum] = i;\r\n      }\r\n    }\r\n    return false;\r\n  }\r\n};<\/pre>\n
    Related Problems<\/strong><\/h1>\n