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Author: Ainevsia

notes/src/SUMMARY.md

@@ -169,4 +169,8 @@
   - [300.最长递增子序列](./day52/lc300.md)
   - [674. 最长连续递增序列](./day52/lc674.md)
   - [718. 最长重复子数组](./day52/lc718.md)
+- [day 53](./day53.md)
+  - [1143.最长公共子序列](./day53/lc1143.md)
+  - [1035.不相交的线](./day53/lc1035.md)
+  - [53. 最大子序和](./day53/lc53.md)
 - [remains](./remains.md)

notes/src/day53.md

@@ -0,0 +1 @@
+1143.最长公共子序列

notes/src/day53/lc1035.md

@@ -0,0 +1,25 @@
+# 1035.不相交的线
+
+原来需要转化成‘最长公共子序列的长度’，一下子真不会
+
+```cpp
+class Solution {
+public:
+    int maxUncrossedLines(vector<int>& v, vector<int>& w) {
+        // dp[i][j]：长度为[0, i - 1]的字符串text1与长度为[0, j - 1]的字符串text2的最长公共子序列为dp[i][j]
+        vector<vector<int>>dp(v.size()+1, vector<int>(w.size()+1, 0));
+        int result = 0;
+        for (int i = 1; i <= v.size(); i++) {
+            for (int j = 1; j <= w.size(); j++) {
+                if (v[i - 1] == w[j - 1]) {
+                    dp[i][j] = dp[i - 1][j - 1] + 1;
+                } else {
+                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
+                }
+                if (dp[i][j] > result) result = dp[i][j];
+            }
+        }
+        return result;
+    }
+};
+```

notes/src/day53/lc1143.md

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+# 1143.最长公共子序列
+
+抄了随想录
+
+```cpp
+class Solution {
+public:
+    int longestCommonSubsequence(string v, string w) {
+        // dp[i][j]：长度为[0, i - 1]的字符串text1与长度为[0, j - 1]的字符串text2的最长公共子序列为dp[i][j]
+        vector<vector<int>>dp(v.size()+1, vector<int>(w.size()+1, 0));
+        int result = 0;
+        for (int i = 1; i <= v.size(); i++) {
+            for (int j = 1; j <= w.size(); j++) {
+                if (v[i - 1] == w[j - 1]) {
+                    dp[i][j] = dp[i - 1][j - 1] + 1;
+                } else {
+                    dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
+                }
+                if (dp[i][j] > result) result = dp[i][j];
+            }
+        }
+        return result;
+    }
+};
+```

notes/src/day53/lc53.md

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+# 53. 最大子序和
+
+自己做的 喜喜
+
+```cpp
+class Solution {
+public:
+    int maxSubArray(vector<int>& v) {
+        // dp[i] 以前i个数结尾的连续子数组最大和
+        int res = v[0];
+        vector<int>dp(v.size(),0);
+        dp[0]=res;
+        for(int i=1;i<v.size();i++){
+            dp[i]=max(v[i],dp[i-1]+v[i]);
+            res=max(res,dp[i]);
+        }
+        return res;
+    }
+};
+```