From 89885c0e530d8b6f51a827d1456f36509f4a1bbe Mon Sep 17 00:00:00 2001 From: markilue <745518019@qq.com> Date: Wed, 15 Mar 2023 11:52:13 +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/T60_MajorityElement.java | 93 ++++++++++++++ .../com/markilue/leecode/hot100/T61_Rob.java | 58 +++++++++ .../leecode/hot100/T62_NumIslands.java | 113 ++++++++++++++++++ .../leecode/hot100/T63_ReverseList.java | 51 ++++++++ 4 files changed, 315 insertions(+) create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/T60_MajorityElement.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/T61_Rob.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/T62_NumIslands.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/T63_ReverseList.java diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/T60_MajorityElement.java b/Leecode/src/main/java/com/markilue/leecode/hot100/T60_MajorityElement.java new file mode 100644 index 0000000..1786ae4 --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/T60_MajorityElement.java @@ -0,0 +1,93 @@ +package com.markilue.leecode.hot100; + +import org.junit.Test; + +import java.util.HashMap; +import java.util.Map; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100 + *@Author: markilue + *@CreateTime: 2023-03-15 10:04 + *@Description: + * TODO 力扣169题 多数元素: + * 给定一个大小为 n 的数组 nums ，返回其中的多数元素。多数元素是指在数组中出现次数 大于 ⌊ n/2 ⌋ 的元素。 + * 你可以假设数组是非空的，并且给定的数组总是存在多数元素。 + *@Version: 1.0 + */ +public class T60_MajorityElement { + + @Test + public void test() { + int[] nums = {3, 2, 3}; + System.out.println(majorityElement(nums)); + } + + + /** + * 思路:按个遍历记录次数 + * 由于数字大小不一定，所以不太好用数组记录，使用map记录 + * 速度击败31.94% 内存击败22.92% 10ms + * @param nums + * @return + */ + public int majorityElement(int[] nums) { + + HashMap map = new HashMap<>();// + int length = nums.length; + + for (int num : nums) { + int count = map.getOrDefault(num, 0) + 1; + if (count > length / 2) return num; + map.put(num, count); + } + + return 0; + + } + + + /** + * 官方最快 + * 速度击败100% 内存击败61.22% + * @param nums + * @return + */ + public int majorityElement1(int[] nums) { + return dfs(nums, 0); + } + + private int dfs(int[] nums, int start) { + int cnt = 1; + for (int i = start + 1; i < nums.length; i++) { + if (nums[i] == nums[start]) cnt++; + else cnt--; + if (cnt == 0) return dfs(nums, i + 1); + } + return nums[start]; + } + + + /** + * 上述思路的正常版 + * 莫斯投票法 + * @param nums + * @return + */ + public int majorityElement2(int[] nums) { + int count = 1; + int maj = nums[0]; + for (int i = 1; i < nums.length; i++) { + if (maj == nums[i]) + count++; + else { + count--;//大于n/2的最后一定不会被减完 + if (count == 0) { + maj = nums[i + 1]; + } + } + } + return maj; + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/T61_Rob.java b/Leecode/src/main/java/com/markilue/leecode/hot100/T61_Rob.java new file mode 100644 index 0000000..cc7426a --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/T61_Rob.java @@ -0,0 +1,58 @@ +package com.markilue.leecode.hot100; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100 + *@Author: markilue + *@CreateTime: 2023-03-15 10:26 + *@Description: + * TODO 力扣198 打家劫舍: + * 你是一个专业的小偷，计划偷窃沿街的房屋。 + * 每间房内都藏有一定的现金，影响你偷窃的唯一制约因素就是相邻的房屋装有相互连通的防盗系统， + * 如果两间相邻的房屋在同一晚上被小偷闯入，系统会自动报警。 + * 给定一个代表每个房屋存放金额的非负整数数组， + * 计算你 不触动警报装置的情况下 ，一夜之内能够偷窃到的最高金额。 + *@Version: 1.0 + */ +public class T61_Rob { + + + /** + * 思路:动态规划法；可以分为当前位置偷；或者当前位置不偷 + * @param nums + * @return + */ + public int rob(int[] nums) { + + int[][] dp = new int[nums.length][2]; + dp[0][0] = 0;//不偷 + dp[0][1] = nums[0];//偷 + + for (int i = 1; i < nums.length; i++) { + dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1]); + dp[i][1] = dp[i - 1][0] + nums[i]; + } + + return Math.max(dp[nums.length - 1][0], dp[nums.length - 1][1]); + + } + + + //滚动数组 + public int rob1(int[] nums) { + + + int dp0 = 0;//不偷 + int dp1 = nums[0];//偷 + int temp; + + for (int i = 1; i < nums.length; i++) { + temp = dp0; + dp0 = Math.max(dp0, dp1); + dp1 = temp + nums[i]; + } + + return Math.max(dp0, dp1); + + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/T62_NumIslands.java b/Leecode/src/main/java/com/markilue/leecode/hot100/T62_NumIslands.java new file mode 100644 index 0000000..b9af97d --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/T62_NumIslands.java @@ -0,0 +1,113 @@ +package com.markilue.leecode.hot100; + +import org.junit.Test; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100 + *@Author: markilue + *@CreateTime: 2023-03-15 10:36 + *@Description: + * TODO 力扣200 岛屿数量: + * 给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。 + * 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。 + * 此外，你可以假设该网格的四条边均被水包围。 + *@Version: 1.0 + */ +public class T62_NumIslands { + + @Test + public void test() { + char[][] grid = { + {'1', '1', '1'}, + {'0', '1', '0'}, + {'1', '1', '1'}, + }; + System.out.println(numIslands(grid)); + } + + + @Test + public void test1() { + char[][] grid = { + {'1', '0', '1', '1', '1'}, + {'1', '0', '1', '0', '1'}, + {'1', '1', '1', '0', '1'}, + }; + System.out.println(numIslands(grid)); + } + + + boolean[][] used; + + /** + * 思路:动态规划？好像不动态规划也行 + * 好像不行 前面的状态可能因为右面的而改变 + * 贪心?碰上相连的就把与他相连的所有都变成 0 + * @param grid + * @return + */ + public int numIslands(char[][] grid) { + + used = new boolean[grid.length][grid[0].length]; + int result = 0; + + for (int i = 0; i < grid.length; i++) { + for (int j = 0; j < grid[0].length; j++) { + if (grid[i][j] == '1') { + clear(grid, i, j); + result++; + } + } + } + + return result; + + + } + + //贪心清空 + public void clear(char[][] grid, int i, int j) { + if (i == grid.length || j == grid[0].length || j < 0 || i < 0) { + return; + } + + if (grid[i][j] == '1' && !used[i][j]) { + used[i][j] = true; + grid[i][j] = '0'; + clear(grid, i + 1, j); + clear(grid, i - 1, j);//注意还要清理上边 + clear(grid, i, j - 1);//注意还要清理左边 + clear(grid, i, j + 1); + } + } + + + + void dfs(char[][] grid, int r, int c) { + if(r < 0 || c < 0 || r >= grid.length || c >= grid[0].length || grid[r][c] == '0'){ + return; + } + grid[r][c] = '0'; + dfs(grid,r + 1,c); + dfs(grid,r - 1,c); + dfs(grid,r,c - 1); + dfs(grid,r,c + 1); + } + //官方最快方法:与本人一致，但是他通过先把当前置为’0‘,可以不用使用used数组 + public int numIslands1(char[][] grid) { + if(grid == null || grid.length == 0){ + return 0; + } + int count = 0; + for(int i = 0;i < grid.length;i++){ + for(int j = 0;j < grid[0].length;j++){ + if(grid[i][j] == '1'){ + count++; + dfs(grid,i,j); + } + } + } + return count; + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/T63_ReverseList.java b/Leecode/src/main/java/com/markilue/leecode/hot100/T63_ReverseList.java new file mode 100644 index 0000000..6d383c4 --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/T63_ReverseList.java @@ -0,0 +1,51 @@ +package com.markilue.leecode.hot100; + +import com.markilue.leecode.listnode.ListNode; +import com.markilue.leecode.listnode.ListNodeUtils; +import org.junit.Test; + +import java.util.List; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100 + *@Author: markilue + *@CreateTime: 2023-03-15 11:29 + *@Description: + * TODO 力扣206题 反转链表: + * 给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。 + *@Version: 1.0 + */ +public class T63_ReverseList { + + @Test + public void test(){ + ListNode root = ListNodeUtils.build(new int[]{1, 2, 3, 4, 5}); + ListNodeUtils.print(reverseList(root)); + } + + + /** + * 思路:使用另一个链表记录 + * 速度击败100% 内存击败33.3% + * @param head + * @return + */ + public ListNode reverseList(ListNode head) { + + ListNode fake = new ListNode(); + + ListNode temp = head; + ListNode tempNext; + + while (temp != null) { + ListNode listNode = new ListNode(temp.val); + listNode.next = fake.next; + fake.next = listNode; + temp=temp.next; + } + + return fake.next; + + } +}