413. Arithmetic Slices

思路:动态规划 参考

解法一:

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int res = 0, len = 2, n = A.size();
        for (int i = 2; i < n; ++i) {
            if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
                ++len;
            } else {
                if (len > 2) res += (len - 1) * (len - 2) * 0.5;
                len = 2;
            }
        }
        if (len > 2) res += (len - 1) * (len - 2) * 0.5;
        return res;
    }
};

解法二:

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        int res = 0, n = A.size();
        vector<int> dp(n, 0);
        for (int i = 2; i < n; ++i) {
            if (A[i] - A[i - 1] == A[i - 1] - A[i - 2]) {
                dp[i] = dp[i - 1] + 1;
            }
            res += dp[i];
        }
        return res;
    }
};
原文地址:https://www.cnblogs.com/bright-mark/p/9537633.html