力扣1027. 最长等差数列

动态规划

• 思路：
  • 可以参考力扣1218. 最长定差子序列
  • 目前不清楚公差，可以将序列最大最小值找到，公差的范围是 -(max - min), (max - min)，按公差递增迭代遍历求出最长等差数列；

class Solution {
public:
    int longestArithSeqLength(vector<int>& nums) {
        auto [minit, maxit] = std::minmax_element(nums.begin(), nums.end());
        int diff = *maxit - *minit;
        int ans = 0;
        for (int d = -diff; d <= diff; ++d) {
            std::unordered_map<int, int> dp;
            for (int v : nums) {
                dp[v] = dp[v - d] + 1;
                ans = std::max(ans, dp[v]);
            }
        }

        return ans;
    }
};

• 时间复杂度比较高，应该是哈希表频繁插入导致，将 dp 数据结构换成数组，数组下标最大值为元素最大值 + 1；

class Solution {
public:
    int longestArithSeqLength(vector<int>& nums) {
        auto [minit, maxit] = std::minmax_element(nums.begin(), nums.end());
        int diff = *maxit - *minit;
        int ans = 1;
        for (int d = -diff; d <= diff; ++d) {
            std::vector<int> dp(*maxit + 1, -1);
            for (int v : nums) {
                int prev = v - d;
                // ensure prev is in nums and has exist(or v is the first item)
                if (prev >= *minit && prev <= *maxit && dp[prev] != -1) {
                    dp[v] = std::max(dp[v], dp[prev] + 1);
                    ans = std::max(ans, dp[v]);
                }
                dp[v] = std::max(dp[v], 1);
            }
        }

        return ans;
    }
};

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