【CODE】Continuous Subarray Sum

523. Continuous Subarray Sum

Medium

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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 a 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:

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

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

• https://blog.csdn.net/u014593748/article/details/70182807
• 使用dp：dp[i][j]=dp[i-1][j]-dp[i-1][i-1];但是数组导致内存不够。
• 上述方法使用公共序列：
• 若n[0]+n[1]+n[2]+...+n[j]=m1*k+q;
• 且n[0]+n[1]+n[2]+...+n[j]+n[j+1]+...+n[p]=m2*k+q;
• 故n[j+1]+...+n[p]=(m2-m1)*k
• 满足条件的话需要 p-(j+1) >=1
• Runtime: 24 ms, faster than 95.54% of C++ online submissions for Continuous Subarray Sum.
• Memory Usage: 11.6 MB, less than 70.00% of C++ online submissions for Continuous Subarray Sum.
• Next challenges:
• Subarray Sum Equals K