leecode，rt_phm更新更新

Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/dfs/LC_1254_ClosedIsland.java

@@ -137,4 +137,33 @@ public class LC_1254_ClosedIsland {
         return dfs(grid, i - 1, j) & dfs(grid, i + 1, j) & dfs(grid, i, j - 1) & dfs(grid, i, j + 1);
     }
 
+
+    //如何为封闭岛屿:有边界
+    public int closedIsland3(int[][] grid) {
+
+        int result = 0;
+        for (int i = 0; i < grid.length; i++) {
+            for (int j = 0; j < grid[0].length; j++) {
+                if (grid[i][j] == 0) {
+                    if (find(grid, i, j)) {
+                        result++;
+                    }
+                }
+            }
+        }
+        return result;
+    }
+
+    public boolean find(int[][] grid, int i, int j) {
+        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length) {
+            return false;//碰到了边界还没有返回,则不是封闭岛屿
+        }
+        if (grid[i][j] == 1 || grid[i][j] == 2) {
+            //遇到了边界
+            return true;
+        }
+        grid[i][j] = 2;
+        return find(grid, i + 1, j) & find(grid, i - 1, j) & find(grid, i, j + 1) & find(grid, i, j - 1);
+    }
+
 }

Leecode/src/main/java/com/markilue/leecode/hot100/second/T49_124_MaxPathSum.java

@@ -99,4 +99,22 @@ public class T49_124_MaxPathSum {
         return Math.max(left, right) + root.val;
 
     }
+
+
+    public int maxPathSum3(TreeNode root) {
+        findCurMax(root);
+        return maxSum;
+    }
+
+    //返回要当前节点的最大值;不要的情况已经在子节点中讨论过了
+    public int findCurMax(TreeNode node) {
+        if (node == null) {
+            return 0;
+        }
+        int left = Math.max(findCurMax(node.left), 0);
+        int right = Math.max(findCurMax(node.right), 0);
+
+        maxSum = Math.max(maxSum, left + right + node.val);
+        return Math.max(left, right) + node.val;
+    }
 }

Leecode/src/main/java/com/markilue/leecode/hot100/second/T67_221_MaximalSquare.java

@@ -20,7 +20,7 @@ public class T67_221_MaximalSquare {
                 {'1', '1', '1', '1', '1'},
                 {'1', '0', '0', '1', '0'}
         };
-        System.out.println(maximalSquare1(matrix));
+        System.out.println(maximalSquare2(matrix));
     }
 
     @Test
@@ -100,4 +100,34 @@ public class T67_221_MaximalSquare {
         return result * result;
 
     }
+
+
+    public int maximalSquare2(char[][] matrix) {
+        int m = matrix.length;
+        int n = matrix[0].length;
+        int[][] dp = new int[m][n];
+        int result = Integer.MIN_VALUE;
+        for (int i = 0; i < n; i++) {
+            dp[0][i] = matrix[0][i] == '1' ? 1 : 0;
+            result = Math.max(result,dp[0][i]);
+        }
+
+        for (int i = 0; i < m; i++) {
+            dp[i][0] = matrix[i][0] == '1' ? 1 : 0;
+            result = Math.max(result,dp[0][i]);
+        }
+
+        for (int i = 1; i < m; i++) {
+            for (int j = 1; j < n; j++) {
+                if (matrix[i][j] == '1') {
+                    dp[i][j] = Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
+                }
+                if (result < dp[i][j]) result = dp[i][j];
+            }
+        }
+        return result * result;
+
+    }
+
+
 }

Leecode/src/main/java/com/markilue/leecode/hot100/second/T69_234_IsPalindrome.java

@@ -33,4 +33,22 @@ public class T69_234_IsPalindrome {
         return false;
 
     }
+
+    public boolean isPalindrome1(ListNode head) {
+        root = head;
+        return find(head);
+    }
+
+    public boolean find(ListNode node) {
+        if (node == null) {
+            return true;
+        }
+
+        if (find(node.next) && node.val == root.val) {
+            root = root.next;
+            return true;
+        }
+
+        return false;
+    }
 }

Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question1.java

@@ -0,0 +1,51 @@
+package com.markilue.leecode.interview.meituan.T0415;
+
+import java.util.Scanner;
+
+/**
+ *@BelongsProject: Leecode
+ *@BelongsPackage: com.markilue.leecode.interview.meituan.T0415
+ *@Author: markilue
+ *@CreateTime: 2023-06-06  11:38
+ *@Description:
+ *  TODO  字符串前缀:
+ *   现在有两个字符串S和T，你需要对S进行若干次操作，使得S是T的一个前缀（空串也是一个前缀）。
+ *   每次操作可以修改S的一个字符，或者删除一个S末尾的字符。小团需要写一段程序，输出最少需要操作的次数。
+ *
+ *@Version: 1.0
+ */
+public class Question1 {
+
+    public static void main(String[] args) {
+        Scanner sc = new Scanner(System.in);
+        int count = sc.nextInt();
+
+        for (int i = 0; i < count; i++) {
+            String S = sc.next();
+            String T = sc.next();
+            solve(S, T);
+        }
+    }
+
+
+    //为什么不是动态规划？因为题目要求，删只能删除S的末尾
+    private static void solve(String s, String t) {
+
+        int result = 0;
+        int pos = s.length() - 1;
+        //如果S是T的前缀，则S一定要比T短
+        if (s.length() > t.length()) {
+            result += s.length() - t.length();
+            pos = t.length() - 1;
+        }
+        //能修改就修改;不能修改再删除
+        for (int i = pos; i >= 0; i--) {
+            if (t.charAt(i) != s.charAt(i)) {
+                result++;
+            }
+        }
+
+
+        System.out.println(result);
+    }
+}