300.最长递增子序列

从前往后遍历即可

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[j] + 1, dp[i]);}if (dp[i] > result) result = dp[i];}return result;}
};

674. 最长连续递增序列

DP

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

贪心

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

718. 最长重复子数组  

nums1和nums2匹配上之后，dp数组状态只能由其左上角推出来

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;}
};