Day.36 | 300.最长递增子序列 674.最长连续递增序列 718.最长重复子数组

300.最长递增子序列

要点：dp[i] 表示在位置 i 处，最长递增子序列的长度

class Solution {
public:int lengthOfLIS(vector<int>& nums) {if (nums.size() <= 1) return nums.size();vector<int> dp(nums.size(), 1);int result = 0;for (int i = 1; i < nums.size(); ++i) {for (int j = 0; j < i; ++j) {if (nums[i] > nums[j])dp[i] = max(dp[i], dp[j] + 1);}result = max(result, dp[i]);}return result;}
};

674.最长连续递增序列

要点：dp[i] = dp[i - 1] + 1;

class Solution {
public:int findLengthOfLCIS(vector<int>& nums) {if (nums.size() <= 1)return nums.size();vector<int> dp(nums.size(), 1);int result = 0;for (int i = 1; i < nums.size(); i++) {if (nums[i] > nums[i - 1]) {dp[i] = dp[i - 1] + 1;} else {dp[i] = 1;}result = max(result, dp[i]);}return result;}
};

718.最长重复子数组

要点：有点难，没有彻底理解 i-1, j-1 的含义

class Solution {
public:int findLength(vector<int>& nums1, vector<int>& nums2) {vector<vector<int>> dp(nums1.size() + 1, vector<int>(nums2.size() + 1, 0));int result = 0;for (int i = 1; i <= nums1.size(); ++i) {for (int j = 1; j <= nums2.size(); ++j) {if (nums1[i - 1] == nums2[j - 1]) {dp[i][j] = dp[i - 1][j - 1] + 1;}result = max(result, dp[i][j]);}}return result;}
};