[LeetCode OJ]Continuous Subarray Sum_ojsubarray sum-优快云博客

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:

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

问题来源：continuous subarray sum

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)。

源代码：

// Solution 1
class Solution {
public:
    bool checkSubarraySum(vector<int>& 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<int>& nums, int k) {
        int sum = 0;
        int pre = 0;
        unordered_set<int> 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;
    }
};