leetcode523. Continuous Subarray Sum

最新推荐文章于 2025-02-16 11:05:14 发布

dike1993

最新推荐文章于 2025-02-16 11:05:14 发布

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分类专栏：算法 leetcode 文章标签： leetcode

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本文介绍了一种算法，用于检查给定数组是否存在长度至少为2的连续子数组，其元素之和为给定整数k的倍数。通过计算前缀和并检查差值是否能被k整除来实现。

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题目

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:

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

“`

思路

可以用遍历求和的方法来做，通过计算nums中i之前的和，然后通过求差值来计算子序列的和，然后判断和是否被k整除，需要考虑k是0的情况，数组大小要大于10000.

代码

 bool checkSubarraySum(vector<int>& nums, int k) {
    if (nums.size() == 0)   
    return false;
    int preSum[10001];
    for (int i = 1; i <= nums.size(); i++) {
        preSum[i] = preSum[i-1] + nums[i-1];
    }

    for (int i = 0; i < nums.size(); i++) {
        for (int j = i+2; j <= nums.size(); j++) {
            if (k == 0) {
                if (preSum[j] - preSum[i] == 0) {
                    return true;
                }
            } else if ((preSum[j] - preSum[i]) % k == 0) {
                return true;
            }
        }
    }
    return false;
}