于是就看了solution。 需要使用一个定理: 代码语言:javascript 代码运行次数:0 publicbooleancheckSubarraySum(int[]nums,int k){Map<Integer,Integer>map=newHashMap<Integer,Integer>();map.put(0,-1);//为了处理nums=[0,0] k=-1这样的情况int sum=0;for(int i=0;i<nums.length;i++){sum+=nums[i...
if(k==0 && sum==0) return true; if(k!=0 && sum%k==0) return true;}} return false; } }; classSolution {public:boolcheckSubarraySum(vector<int>& nums,intk) {intsum=0; map<int,int>sumhash;for(inti=0;i<nums.size();i++){ sum+=nums[i];//judge if k=0if(k==0&& i<...
https://leetcode.com/problems/continuous-subarray-sum/discuss/99567/java-solution https://leetcode.com/problems/continuous-subarray-sum/discuss/99499/java-on-time-ok-space https://leetcode.com/problems/continuous-subarray-sum/discuss/99506/concise-c-solution-use-set-instead-of-map LeetCode All in...
Can you solve this real interview question? Continuous Subarray Sum - Given an integer array nums and an integer k, return true if nums has a good subarray or false otherwise. A good subarray is a subarray where: * its length is at least two, and * t
523. Continuous Subarray Sum 题意:计算是否存在和为k的倍数连续子数组。 思路:不是很懂,大致上是说,如果第一次遇见不存在的模的结果时,则保存下来,但是如果遇到多个相同的求模的结果,说明它们之间的和必然是k的倍数。 class Solution(object): def
1classSolution {2func checkSubarraySum(_ nums: [Int], _ k: Int) ->Bool {3ifnums.count <2{returnfalse}4if(k ==0) {returnnums[0] ==0&& nums[1] ==0}5vardict =[Int:Int]()67varsum =08foriin0...nums.count-1{9sum +=nums[i]1011let r = sum %k12ifr ==0{1314if(i>=1...
LeetCode 523: Continuous Subarray Sum Note: 1. The sum array need to be very clear that 0th is 0. So the sum[i] means from 0 to i - 1 sum. classSolution {publicbooleancheckSubarraySum(int[] nums,intk) {if(nums.length < 2) {returnfalse;...
【leetcode】523. Continuous Subarray Sum 题目如下: 解题思路:本题需要用到这么一个数学定理。对于任意三个整数a,b,k(k !=0),如果 a%k = b%k,那么(a-b)%k = 0。利用这个定理,我们可以对数组从头开始进行求和,同时利用字典保存余数(key:余数,value:最早出现这个余数的元素下标),每累加一个元素都对k...
Leetcode 523. Continuous Subarray Sum prefix sum classSolution(object):defcheckSubarraySum(self, nums, k):""":type nums: List[int] :type k: int :rtype: bool"""ifk ==0:returnany(nums[i - 1] == 0andnums[i] == 0foriinrange(1, len(nums)))...
classSolution {public:boolcheckSubarraySum(vector<int>& nums,intk) { unordered_map<int,int>mp;intsum =0; mp[0] = -1;for(inti =0; i < nums.size(); ++i) { sum+=nums[i];if(k) sum %=k;if(mp.count(sum) >0) {if(i - mp[sum] >1)returntrue; ...