Leetcode0523--Continuous Subarray Sum 連續和倍數

 【轉載請註明】http://www.javashuo.com/article/p-nkvbhhnp-cg.html

 

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

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        int sum=0;
        map<int,int> sumhash;
        for(int i=0;i<nums.size();i++){
            sum+=nums[i];
            // judge if k=0
            if(k==0 && i<nums.size()-1){
                if((nums[i]+nums[i+1])==0) return true;
            }
            // have previous sum%k
            if(k!=0 && sumhash.find(sum%k)!=sumhash.end()) return true;
            if(i!=0 && k!=0 && sum%k==0) return true;
            if(k!=0) sumhash[sum%k]++;
        }
        return false;
    }
};

 int t = (k == 0) ? sum : (sum % k);
 if (m.count(t)) {
    if (i - m[t] > 1) return true;
 } else m[t] = i;