300.最长递增子序列
参考
- dp[i] 表示以 i 为结尾的最长递增子序列长度
- 递推公式: 使用 i 和 j 判断
- dp[i] = max(dp[j] + 1, dp[i])
- 每次 j 都需要从头遍历
- 初始化: dp[i] = 1
class Solution {
public:int lengthOfLIS(vector<int>& nums) {vector<int> dp(nums.size(), 1);for (int i = 1; i < nums.size(); i++) {for (int j = 0; j < i; j++) {if (nums[j] < nums[i]) {dp[i] = max(dp[j] + 1, dp[i]);}}}int res = 0;for (int i = 0; i < nums.size(); i++) {res = max(res, dp[i]);}return res;}
};
674. 最长连续递增序列
区别在于是否连续, 无需重新开始遍历
class Solution {
public:int findLengthOfLCIS(vector<int>& nums) {vector<int> dp(nums.size(), 1);for (int i = 1; i < nums.size(); i++) {if (nums[i - 1] < nums[i]) {dp[i] = dp[i - 1] + 1;}}int res = 0;for (int i = 0; i < dp.size(); i++) {res = max(res, dp[i]);}return res;}
};
718. 最长重复子数组
参考
dp[i][j] :以下标i - 1为结尾的A,和以下标j - 1为结尾的B,最长重复子数组长度为dp[i][j]
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 res = -1;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;res = max(res, dp[i][j]);}}return res;}
};