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[LeetCode OJ]Continuous Subarray Sum

编程语言 sysu_xiamengyou 10℃ 0评论
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1.问题描述:

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=6
Output: True
Explanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.

Example 2:

Input: [23, 2, 6, 4, 7],  k=6
Output: True
Explanation: 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



2.解题分析:

这是一个用动态规划思路来解决的问题。




方法一:遍历数组中的每一个数,尝试分别以他们作为连续子数组的起点,然后遍历这些起点以后的数字,找出符合条件的子数组;时间复杂度为遍历两次数组空间,为O(N^2)。



方法二:如果a % k = c,b % k = c,那么(a – b) % k = 0。根据这一定理,那么我们计算从数组起点开始的和,利用set集合来存这个和除以k的余值(k≠0,如果k=0,则存这个和的值),如果当前的累加和除以k的余值已经在set集合中存在了,那么数组空间中必有一段子数组的和是k或者k的整数倍。由于题意规定至少是两个数的子数组,需要一个pre来记录之前的余值。时间复杂度为遍历一次数组空间,为O(N)。



3.源代码:

// Solution 1
class Solution {
public:
    bool checkSubarraySum(vector& nums, int k) {
        for (int i = 0; i < nums.size(); i++) {
         int sum = nums[i];
         for (int j = i + 1; j < nums.size(); j++) {
          sum += nums[j];
          if (sum == k) {
           return true;
          }
          if (k != 0 && sum % k == 0) {
           return true;
          }
         }
        }
        return false;
    }
};

// Solution 2
class Solution {
public:
    bool checkSubarraySum(vector& nums, int k) {
        int sum = 0;
        int pre = 0;
        unordered_set res;
        for (int i = 0; i < nums.size(); i++) {
         sum += nums[i];
         int t = (k == 0) ? sum : (sum % k);
         if (res.count(t)) {
          return true;
         }
         res.insert(pre);
         pre = t;
        }
        return false;
    }
};





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