From 33c16119c62a44d151ea279b3080fa914ac9a801 Mon Sep 17 00:00:00 2001 From: markilue <745518019@qq.com> Date: Wed, 26 Apr 2023 13:12:09 +0800 Subject: [PATCH] =?UTF-8?q?leecode=E6=9B=B4=E6=96=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../leecode/hot100/second/T23_42_Trap.java | 4 +- .../leecode/hot100/second/T24_46_Permute.java | 54 +++++++++++++ .../leecode/hot100/second/T25_48_Rotate.java | 43 ++++++++++ .../hot100/second/T27_53_MaxSubArray.java | 14 ++++ .../leecode/hot100/second/T28_55_CanJump.java | 29 +++++++ .../leecode/hot100/second/T29_56_Merge.java | 23 ++++++ .../hot100/second/T30_62_UniquePaths.java | 59 ++++++++++++++ .../hot100/second/T31_64_MinPathSum.java | 81 +++++++++++++++++++ 8 files changed, 305 insertions(+), 2 deletions(-) create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/second/T24_46_Permute.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/second/T25_48_Rotate.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/second/T28_55_CanJump.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/second/T30_62_UniquePaths.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/second/T31_64_MinPathSum.java diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T23_42_Trap.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T23_42_Trap.java index 644d192..1c30c60 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T23_42_Trap.java +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T23_42_Trap.java @@ -41,10 +41,10 @@ public class T23_42_Trap { if (height[left] < height[right]) { //左边一定的积水 - sum += Math.min(leftMax, rightMax) - height[left]; + sum += leftMax - height[left]; left++; } else { - sum += Math.min(leftMax, rightMax) - height[right]; + sum += rightMax - height[right]; right--; } diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T24_46_Permute.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T24_46_Permute.java new file mode 100644 index 0000000..adcae32 --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T24_46_Permute.java @@ -0,0 +1,54 @@ +package com.markilue.leecode.hot100.second; + +import java.util.ArrayList; +import java.util.Arrays; +import java.util.List; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.second + *@Author: markilue + *@CreateTime: 2023-04-26 10:25 + *@Description: + * TODO 力扣46 全排列: + * 给定一个不含重复数字的数组 nums ，返回其 所有可能的全排列 。你可以 按任意顺序 返回答案。 + *@Version: 1.0 + */ +public class T24_46_Permute { + + List> result = new ArrayList<>(); + List cur = new ArrayList<>(); + + + /** + * 重点需要做树枝去重 + * @param nums + * @return + */ + public List> permute(int[] nums) { + Arrays.sort(nums); + backtracking(nums, new boolean[nums.length]); + + return result; + } + + public void backtracking(int[] nums, boolean[] used) { + if (nums.length == cur.size()) { + result.add(new ArrayList<>(cur)); + return; + } + + for (int i = 0; i < nums.length; i++) { + if (used[i]) continue; + if (i > 0 && nums[i] == nums[i - 1]) { + continue; + } + used[i] = true; + cur.add(nums[i]); + backtracking(nums, used); + cur.remove(cur.size() - 1); + used[i] = false; + } + + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T25_48_Rotate.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T25_48_Rotate.java new file mode 100644 index 0000000..bc0fb2d --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T25_48_Rotate.java @@ -0,0 +1,43 @@ +package com.markilue.leecode.hot100.second; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.second + *@Author: markilue + *@CreateTime: 2023-04-26 10:39 + *@Description: + * TODO 力扣48 旋转图像: + * 给定一个 n × n 的二维矩阵 matrix 表示一个图像。请你将图像顺时针旋转 90 度。 + * 你必须在 原地 旋转图像，这意味着你需要直接修改输入的二维矩阵。请不要 使用另一个矩阵来旋转图像。 + *@Version: 1.0 + */ +public class T25_48_Rotate { + + + /** + * 官方题解中的一种方法: + * 先水平反转,在对角线先反转 + * @param matrix + */ + public void rotate(int[][] matrix) { + + //水平反转 + for (int i = 0; i < matrix.length / 2; i++) { + for (int j = 0; j < matrix[0].length; j++) { + int temp = matrix[i][j]; + matrix[i][j] = matrix[matrix.length - 1 - i][j]; + matrix[matrix.length - 1 - i][j] = temp; + } + } + + //对角线反转 + for (int i = 0; i < matrix.length; i++) { + for (int j = 0; j < i; j++) { + int temp = matrix[i][j]; + matrix[i][j] = matrix[j][i]; + matrix[j][i] = temp; + } + } + + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T27_53_MaxSubArray.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T27_53_MaxSubArray.java index 5b80e1e..8e4e00d 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T27_53_MaxSubArray.java +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T27_53_MaxSubArray.java @@ -58,4 +58,18 @@ public class T27_53_MaxSubArray { } + + + public int maxSubArray2(int[] nums) { + + int dp0=Integer.MIN_VALUE; + int dp1=nums[0]; + + for(int i=1;i maxIndex) return false; + maxIndex = Math.max(maxIndex, i + nums[i]); + if (maxIndex >= nums.length - 1) return true; + } + return maxIndex >= nums.length - 1; + + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T29_56_Merge.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T29_56_Merge.java index 2af2205..e1f9e14 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T29_56_Merge.java +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T29_56_Merge.java @@ -53,4 +53,27 @@ public class T29_56_Merge { return result.toArray(new int[0][]); } + + + public int[][] merge1(int[][] intervals) { + Arrays.sort(intervals, ((o1, o2) -> { + return o1[0] == o2[0] ? o1[1] - o2[1] : o1[0] - o2[0]; + })); + ArrayList result = new ArrayList<>(); + result.add(intervals[0]); + + for (int i = 1; i < intervals.length; i++) { + if (intervals[i][0] <= result.get(result.size() - 1)[1]) { + int[] pre = result.get(result.size() - 1); + pre[1] = Math.max(intervals[i][1], pre[1]); + } else { + result.add(intervals[i]); + } + } + + + return result.toArray(new int[0][]); + } + + } diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T30_62_UniquePaths.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T30_62_UniquePaths.java new file mode 100644 index 0000000..05b932f --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T30_62_UniquePaths.java @@ -0,0 +1,59 @@ +package com.markilue.leecode.hot100.second; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.second + *@Author: markilue + *@CreateTime: 2023-04-26 11:26 + *@Description: + * TODO 力扣62 不同路径: + * 一个机器人位于一个 m x n 网格的左上角 （起始点在下图中标记为 “Start” ）。 + * 机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角（在下图中标记为 “Finish” ）。 + * 问总共有多少条不同的路径？ + *@Version: 1.0 + */ +public class T30_62_UniquePaths { + + public int uniquePaths(int m, int n) { + + int[][] dp = new int[m][n]; + + for (int i = 0; i < m; i++) { + dp[i][0] = 1; + } + + for (int i = 0; i < n; i++) { + dp[0][i] = 1; + } + + for (int i = 1; i < m; i++) { + for (int j = 1; j < n; j++) { + dp[i][j] = dp[i - 1][j] + dp[i][j - 1]; + } + } + + return dp[m - 1][n - 1]; + + } + + + //滚动数组优化 + public int uniquePaths1(int m, int n) { + + int[] dp = new int[n]; + + + for (int i = 0; i < n; i++) { + dp[i] = 1; + } + + for (int i = 1; i < m; i++) { + for (int j = 1; j < n ; j++) { + dp[j] = dp[j] + dp[j - 1]; + } + } + + return dp[n - 1]; + + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/second/T31_64_MinPathSum.java b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T31_64_MinPathSum.java new file mode 100644 index 0000000..8d615ae --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/second/T31_64_MinPathSum.java @@ -0,0 +1,81 @@ +package com.markilue.leecode.hot100.second; + +import org.junit.Test; + +import java.util.Arrays; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.second + *@Author: markilue + *@CreateTime: 2023-04-26 11:35 + *@Description: + * TODO 力扣64 最小路径和: + * 给定一个包含非负整数的 m x n 网格 grid ，请找出一条从左上角到右下角的路径，使得路径上的数字总和为最小。 + * 说明：每次只能向下或者向右移动一步。 + *@Version: 1.0 + */ +public class T31_64_MinPathSum { + + @Test + public void test() { + +// int[][] grid = {{1, 3, 1}, {1, 5, 1}, {4, 2, 1}}; + int[][] grid = {{1, 2, 3}, {4, 5, 6}}; + System.out.println(minPathSum(grid)); + + + } + + public int minPathSum(int[][] grid) { + + int[][] dp = new int[grid.length][grid[0].length]; + + dp[0][0] = grid[0][0]; + + for (int i = 1; i < grid.length; i++) { + dp[i][0] = dp[i - 1][0] + grid[i][0]; + } + + for (int i = 1; i < grid[0].length; i++) { + dp[0][i] = dp[0][i - 1] + grid[0][i]; + } + + for (int i = 1; i < grid.length; i++) { + for (int j = 1; j < grid[0].length; j++) { + dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j]; + } + } + + return dp[grid.length - 1][grid[0].length - 1]; + } + + + int[][] memo; + + //回溯+记忆化搜索 + public int minPathSum1(int[][] grid) { + memo = new int[grid.length][grid[0].length]; + for (int i = 0; i < memo.length; i++) { + Arrays.fill(memo[i], -1); + } + + return backtracking(grid, grid.length - 1, grid[0].length-1); + } + + public int backtracking(int[][] grid, int i, int j) { + if (i < 0 || j < 0) { + return Integer.MAX_VALUE; + } + if (i == 0 && j == 0) { + return grid[0][0]; + } + if (memo[i][j] != -1) { + return memo[i][j]; + } else { + memo[i][j] = Math.min(backtracking(grid, i - 1, j), backtracking(grid, i, j - 1)) + grid[i][j]; + } + + return memo[i][j]; + } +}