From 4d086839527b74f17a49a28b19145dc186149f0b Mon Sep 17 00:00:00 2001 From: kevinding1125 <745518019@qq.com> Date: Tue, 6 Jun 2023 11:54:00 +0800 Subject: [PATCH] =?UTF-8?q?leecode=EF=BC=8Crt=5Fphm=E6=9B=B4=E6=96=B0?= =?UTF-8?q?=E6=9B=B4=E6=96=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../dfs/LC_1254_ClosedIsland.java | 29 +++++++++++ .../hot100/second/T49_124_MaxPathSum.java | 18 +++++++ .../hot100/second/T67_221_MaximalSquare.java | 32 +++++++++++- .../hot100/second/T69_234_IsPalindrome.java | 18 +++++++ .../interview/meituan/T0415/Question1.java | 51 +++++++++++++++++++ 5 files changed, 147 insertions(+), 1 deletion(-) create mode 100644 Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question1.java diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/dfs/LC_1254_ClosedIsland.java b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/dfs/LC_1254_ClosedIsland.java index 88a663b..32b65e4 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/dfs/LC_1254_ClosedIsland.java +++ b/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); + } + } diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T49_124_MaxPathSum.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T49_124_MaxPathSum.java index 2f7e356..73dd7aa 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T49_124_MaxPathSum.java +++ b/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; + } } diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T67_221_MaximalSquare.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T67_221_MaximalSquare.java index 33dc639..90f2c81 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T67_221_MaximalSquare.java +++ b/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; + + } + + } diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T69_234_IsPalindrome.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T69_234_IsPalindrome.java index aa75929..2a82a54 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T69_234_IsPalindrome.java +++ b/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; + } } diff --git a/Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question1.java b/Leecode/src/main/java/com/markilue/leecode/interview/meituan/T0415/Question1.java new file mode 100644 index 0000000..cba4bcb --- /dev/null +++ b/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); + } +}