Continuous Subarray Sum
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本次題目是有關於動態規劃的應用,Continuous Subarray Sum,題目的大致要求如下:
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=6Output: TrueExplanation: Because [2, 4] is a continuous subarray of size 2 and sums up to 6.
Example 2:
Input: [23, 2, 6, 4, 7], k=6Output: TrueExplanation: Because [23, 2, 6, 4, 7] is an continuous subarray of size 5 and sums up to 42.對於求解的過程,我的思路大致如下,首先考慮給定數列n,並構造其各階級數
S(n) = a(1) + a(2) + ... + a(n), n ≥ 1,
進一步可以考慮遞歸算法
S(n) = a(1), n = 1
S(n) = a(n) + S(n - 1), n > 1,
考慮到子數列的和可以寫成這樣的形式
S(n) = a(1), n = 1
S(n) = a(n) + S(n - 1), n > 1。
有了這樣的構思就方便我們去設計此代碼
bool checkSubarraySum(vector<int>& nums, int k)
{
map<int, int> sum; //構造各階級數
for(int i = 0; i < nums.size(); i++)
{
sum[i] = sum[i - 1] + nums[i];
}
for(int i = 0; i < nums.size(); i++)
{
for(int j = i + 1; j < nums.size(); j++)
{
if (k != 0 && (sum[j] - sum[i - 1]) % k == 0 || k == 0 && (sum[j] - sum[i - 1]) == 0)
{
return true;
}
}
}
return false;
}
{
map<int, int> sum; //構造各階級數
for(int i = 0; i < nums.size(); i++)
{
sum[i] = sum[i - 1] + nums[i];
}
for(int i = 0; i < nums.size(); i++)
{
for(int j = i + 1; j < nums.size(); j++)
{
if (k != 0 && (sum[j] - sum[i - 1]) % k == 0 || k == 0 && (sum[j] - sum[i - 1]) == 0)
{
return true;
}
}
}
return false;
}
應用到動態規劃的概念,if (k != 0 && (sum[j] - sum[i - 1]) % k == 0 || k == 0 && (sum[j] - sum[i - 1]) == 0) 判斷每一種不同子數列相加或n*k(mode k)的可能。
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