- dp[i]表示以i为结尾的子序列的最长长度。
- 递推公式：dp[i]=max(dp[i], dp[j]+1)
- 初始化为1
- 遍历顺序：从前到后
class Solution {
public:int lengthOfLIS(vector<int>& nums) {vector<int> dp(nums.size(), 1);int result = 1;for (int i = 1; i < dp.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;}
};
class Solution {
public:int findLengthOfLCIS(vector<int>& nums) {int length = nums.size();vector<int> dp(length, 1);int result = 1;for (int i = 1; i < length; ++i) {if (nums[i] > nums[i - 1])dp[i] = dp[i - 1] + 1;result = max(result, dp[i]);}return result;}
};
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;}if (dp[i][j] > result)result = dp[i][j];}}return result;}
};
