From 1df4dce36e2a9a688b036b461e08a41cf2e8e34e Mon Sep 17 00:00:00 2001 From: kevinding1125 <745518019@qq.com> Date: Mon, 29 May 2023 13:04:32 +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 | 30 ++++++- .../graph/LC_127_LadderLength.java | 62 ++++++++++++++ .../LC_84_LargestRectangleArea.java | 51 ++++++++++++ .../LC_684_FindRedundantConnection.java | 82 +++++++++++++++++++ .../LC_685_FindRedundantConnection.java | 22 +++++ 5 files changed, 246 insertions(+), 1 deletion(-) create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/singlestack/LC_84_LargestRectangleArea.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/LC_684_FindRedundantConnection.java create mode 100644 Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/LC_685_FindRedundantConnection.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 fed8db5..88a663b 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 @@ -21,7 +21,7 @@ public class LC_1254_ClosedIsland { {1, 0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 1, 1, 1, 0} }; - System.out.println(closedIsland(grid)); + System.out.println(closedIsland2(grid)); } @Test @@ -109,4 +109,32 @@ public class LC_1254_ClosedIsland { //&&是短路与，只要有一个不满足就不会判断后面的;&是逻辑与,会判断完后面的 return dfs1(grid, i, j + 1) & dfs1(grid, i, j - 1) & dfs1(grid, i + 1, j) & dfs1(grid, i - 1, j); } + + + //思路:其实本质上就是在岛屿数量前面加上了封闭条件 + //一个岛屿是否封闭:就是看他是否抵达边界,如果抵达边界则不封闭,反之则封闭 + public int closedIsland2(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 && dfs(grid, i, j)) { + result++; + } + } + } + return result; + } + + public boolean dfs(int[][] grid, int i, int j) { + if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length) { + return false;//抵达边界了，不行 + } + if (grid[i][j] == 1) { + return true;//遇上水 + } + grid[i][j] = 1; + //四个方向都要是；同时不能只判断一个，所以使用&而不是&& + return dfs(grid, i - 1, j) & dfs(grid, i + 1, j) & dfs(grid, i, j - 1) & dfs(grid, i, j + 1); + } + } diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/graph/LC_127_LadderLength.java b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/graph/LC_127_LadderLength.java index 30a2bdd..ab4ec90 100644 --- a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/graph/LC_127_LadderLength.java +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/graph/LC_127_LadderLength.java @@ -1,5 +1,7 @@ package com.markilue.leecode.hot100.interviewHot.graph; +import com.sun.crypto.provider.PBEWithMD5AndDESCipher; + import java.util.*; /** @@ -181,4 +183,64 @@ public class LC_127_LadderLength { } return distance < 2; } + + + public int ladderLength3(String beginWord, String endWord, List wordList) { + + //TODO 1.放入HashSet方便判断 + HashSet wordSet = new HashSet<>(wordList); + if (wordSet.size() == 0 || !wordSet.contains(endWord)) { + return 0;//到不了 + } + wordSet.remove(beginWord); + + //TODO 2.创建一个queue存放对应的word + HashSet visited = new HashSet<>(); + ArrayDeque queue = new ArrayDeque<>(); + queue.push(beginWord); + int step = 1;//最begin开始，至少走一步 + while (!queue.isEmpty()) { + int curSize = queue.size(); + for (int i = 0; i < curSize; i++) { + String curWord = queue.pollFirst(); + //任意改变一个字母，判断在不在HashSet里面或者是不是endWord + if (changeEveryChar(curWord, endWord, wordSet, visited, queue)) { + return step + 1; + } + } + step++; + } + + return 0; + + } + + private boolean changeEveryChar(String curWord, String endWord, Set wordSet, Set visited, Deque queue) { + char[] charArray = curWord.toCharArray(); + + for (int i = 0; i < charArray.length; i++) { + char originalLetter = charArray[i]; + //变换成26个字母,判断在不在wordSet里 + for (char j = 'a'; j <= 'z'; j++) { + if (j == originalLetter) continue; + charArray[i] = j; + String changeWord = String.valueOf(charArray); + + //wordSet里面包不包含 + if (wordSet.contains(changeWord)) { + if (endWord.equals(changeWord)) { + return true;//抵达终点 + } + if (!visited.contains(changeWord)) { + queue.addLast(changeWord); + visited.add(changeWord); + } + } + } + charArray[i] = originalLetter; + } + return false; + } + + } diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/singlestack/LC_84_LargestRectangleArea.java b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/singlestack/LC_84_LargestRectangleArea.java new file mode 100644 index 0000000..903c93d --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/singlestack/LC_84_LargestRectangleArea.java @@ -0,0 +1,51 @@ +package com.markilue.leecode.hot100.interviewHot.singlestack; + +import java.util.ArrayDeque; +import java.util.Deque; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.interviewHot.singlestack + *@Author: markilue + *@CreateTime: 2023-05-29 09:46 + *@Description: TODO 力扣84 柱状图中最大的矩形 + *@Version: 1.0 + */ +public class LC_84_LargestRectangleArea { + + //为什么需要单调栈:其实本质上，当前位置的面最大值依赖于左右两边的宽度最大可以延展到哪里 + //而当前位置需要延展到两边,所以这里使用单调栈来记录 + public int largestRectangleArea(int[] heights) { + + Deque stack = new ArrayDeque<>(); + int n = heights.length; + + int[] left = new int[n]; + int[] right = new int[n]; + + for (int i = 0; i < n; i++) { + while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {//寻找第一个比当前值小的值 + stack.pop(); + } + left[i] = stack.isEmpty() ? -1 : stack.peek(); + stack.push(i); + } + stack.clear(); + + for (int i = n - 1; i >= 0; i--) { + while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) { + stack.pop(); + } + right[i] = stack.isEmpty() ? n : stack.peek(); + stack.push(i); + } + + //计算每一个位置的面积,求最大值 + int result = 0; + for (int i = 0; i < n; i++) { + result = Math.max(result, heights[i] * (right[i] - left[i] - 1)); + } + + return result; + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/LC_684_FindRedundantConnection.java b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/LC_684_FindRedundantConnection.java new file mode 100644 index 0000000..f8cdfeb --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/LC_684_FindRedundantConnection.java @@ -0,0 +1,82 @@ +package com.markilue.leecode.hot100.interviewHot.union_find; + +import org.junit.Test; + +import java.util.Arrays; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.interviewHot.union_find + *@Author: markilue + *@CreateTime: 2023-05-29 11:00 + *@Description: TODO 力扣684 冗余连接: + *@Version: 1.0 + */ +public class LC_684_FindRedundantConnection { + + @Test + public void test() { + int[][] edges = { + {1, 4}, + {3, 4}, + {1, 3}, + {1, 2}, + {4, 5}, + }; + System.out.println(Arrays.toString(findRedundantConnection(edges))); + } + + int n;//节点的数量 + int[] father;//父节点矩阵，记录每个子节点的父节点 + + //初始化父节点矩阵 + private void init(int[] father) { + for (int i = 0; i < father.length; i++) { + father[i] = i; + } + } + + //寻找一个节点的父节点是谁 + private int find(int u) { + if (u == father[u]) { + return u; + } + father[u] = find(father[u]); + return father[u]; + } + + //将两个节点 并起来 + private void join(int u, int v) { + u = find(u);//这里是核心 + v = find(v);//这里是核心 + if (u == v) return;//已经在一个集合里面了 + //不是找v的father是u，而是v的father是u的father + father[v] = u;//合并在一起 + } + + //判断两个节点在不在一个并查集里面 + private boolean same(int u, int v) { + int fatherU = find(u); + int fatherV = find(v); + return fatherU == fatherV;//已经在一个集合里面了 + } + + + //本质上就是判断是不是有环，如果有环应该拆哪一个:使用并查集实现 + public int[] findRedundantConnection(int[][] edges) { + n = 1005; + father = new int[n]; + init(father); + for (int[] edge : edges) { + if (same(edge[0], edge[1])) { + //在一个并查集里面 + return edge; + } else { + //不在一个并查集里面,那么就把他们两加入，并连起来 + join(edge[0], edge[1]); + } + } + + return null; + } +} diff --git a/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/LC_685_FindRedundantConnection.java b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/LC_685_FindRedundantConnection.java new file mode 100644 index 0000000..c73a060 --- /dev/null +++ b/Leecode/src/main/java/com/markilue/leecode/hot100/interviewHot/union_find/LC_685_FindRedundantConnection.java @@ -0,0 +1,22 @@ +package com.markilue.leecode.hot100.interviewHot.union_find; + +import org.junit.Test; + +import java.util.Arrays; + +/** + *@BelongsProject: Leecode + *@BelongsPackage: com.markilue.leecode.hot100.interviewHot.union_find + *@Author: markilue + *@CreateTime: 2023-05-29 11:00 + *@Description: TODO 力扣685 冗余连接II: + *@Version: 1.0 + */ +public class LC_685_FindRedundantConnection { + + public int[] findRedundantDirectedConnection(int[][] edges) { + return null; + } + + +}