LN : leetcode 413 Arithmetic Slices
lc 413 Arithmetic Slices
A sequence of number is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.
For example, these are arithmetic sequence:
1, 3, 5, 7, 9
7, 7, 7, 7
3, -1, -5, -9
The following sequence is not arithmetic.
1, 1, 2, 5, 7
A zero-indexed array A consisting of N numbers is given. A slice of that array is any pair of integers (P, Q) such that 0 <= P < Q < N.
A slice (P, Q) of array A is called arithmetic if the sequence:
A[P], A[p + 1], ..., A[Q - 1], A[Q] is arithmetic. In particular, this means that P + 1 < Q.
The function should return the number of arithmetic slices in the array A.
Example:
A = [1, 2, 3, 4]
return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] itself.
规律公式 Accepted
这个问题就是找有几个长度大于等于3的等差数列，我们知道如果有一个等差数列长为n，可将其裂为(n-1)*(n-2)/2个长度大于等于3的子数列。
class Solution {
public:
int numberOfArithmeticSlices(vector<int>& A) {
int len = 2, sum = 0;
for (int i = 2; i < A.size(); i++) {
if (A[i]+A[i-2] == 2*A[i-1]) {
len++;
} else {
if (len >= 3) sum += (len-1)*(len-2)/2;
len = 2;
}
}
if (len >= 3) sum += (len-1)*(len-2)/2;
return sum;
}
};
DP Accepted
如果用dp[i]表示到i位置为止的算数切片的个数，若能与之前两个数构成等差数列，则dp[i] = dp[i-1]+1，累加所有的dp[i]即可得到答案。
class Solution {
public:
int numberOfArithmeticSlices(vector<int>& A) {
vector<int> dp(A.size(), 0);
int ans = 0;
for (int i = 2; i < A.size(); i++) {
if (A[i]+A[i-2] == 2*A[i-1]) dp[i] = dp[i-1]+1;
ans += dp[i];
}
return ans;
}
};
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