Day.38 | 1143.最长公共子序列 1035.不相交的线 53.最大子序和 392.判断子序列

1143.最长公共子序列

要点：dp[i][j] = dp[i - 1][j - 1] + 1;  dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);

class Solution {
public:int longestCommonSubsequence(string text1, string text2) {vector<vector<int>> dp(text1.size() + 1, vector<int>(text2.size() + 1, 0));for (int i = 1; i <= text1.size(); ++i) {for (int j = 1; j <= text2.size(); ++j) {if (text1[i - 1] == text2[j - 1]) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);}}}return dp[text1.size()][text2.size()];}
};

1035.不相交的线

要点：与上一题一模一样

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

53.最大子序和

要点：dp[i] = max(dp[i - 1] + nums[i], nums[i]); 求最大的连续子数组的和，只有两个状态，要么加上当前的 nums[i]，要么从 nums[i] 重新开始

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

392.判断子序列

要点：双指针也能做，但是主要目的是用dp解决，依然与上面几道题目类似

class Solution {
public:bool isSubsequence(string s, string t) {vector<vector<int>> dp(s.size() + 1, vector<int>(t.size() + 1, 0));for (int i = 1; i <= s.size(); ++i) {for (int j = 1; j <= t.size(); ++j) {if (s[i - 1] == t[j - 1]) {dp[i][j] = dp[i - 1][j - 1] + 1;} else {dp[i][j] = dp[i][j - 1];}}}if (dp[s.size()][t.size()] == s.size()) {return true;} else {return false;}}
};