LeetCode 题解(Week 8): 523. Continuous Subarray Sum
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原题目:
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.
中文大意:
给定一个数组nums,其中元素都大于等于0,以及一个整数k,判断在数组中是否存在一个长度大于2,连续的子数组,这个子数组中的数相加是k的倍数
题解:
class Solution {public: bool checkSubarraySum(vector<int>& nums, int k) { bool res = false; int n = nums.size(); if(n<2) return res; for (int i = 0; i < n-1 ; i++) { vector<int> v(n,0); v[i] = nums[i]; for(int j = i+1 ; j < n ;j++) { v[j] = v[j-1] + nums[j]; if((k!=0 && v[j]%k==0) || (k==0 && v[j] ==0)) { res = true; return res; } } } return res; }};
思路:
水题,只需要通过累加的方式算出任意两个下标m,n(n>m)中的和,然后再判断这个和是否是k的倍数(如果k大于0,就判断模k是否等于0,如果k等于0,就判断这个和是否等于0)
算法的时间复杂度为O(n^2),空间复杂度为O(n),因为我用了中间变量来存累加的结果
题解2
在论坛中看到了十分奇妙的思路与题解,分享之。
class Solution {public: bool checkSubarraySum(vector<int>& nums, int k) { int n = nums.size(), sum = 0, pre = 0; unordered_set<int> modk; for (int i = 0; i < n; ++i) { sum += nums[i]; int mod = k == 0 ? sum : sum % k; if (modk.count(mod)) return true; modk.insert(pre); pre = mod; } return false; }};
思路
在这个题解中,通过一个集合来记录上一次的累加模K的结果,当当前的结果(假设下标为i)和与之前结果相同(假设下标为j),说明(i,j)之间的数,相加起来,模K刚好为0,这个时候我们就找到了一组解了!
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