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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