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一、交叉链表

问题：

给你两个单链表的头节点 headA 和 headB ，请你找出并返回两个单链表相交的起始节点。如果两个链表不存在相交节点，返回 null 。

解题思想：

双指针

备注：不是快慢指针，如果两个长度相同可以用快慢指针，因为两个链表长度不同。

package org.example.YangQianGuan;class Node {int data;Node next;Node(int val) {this.data = val;this.next = null;}@Overridepublic String toString() {return " {data: " + this.data + " next: " + this.next + "} ";}
}public class IntersectionNode {public Node getStartInter(Node headA, Node headB) {if (headA == null || headB == null) {return null;}Node pA = headA;Node pB = headB;while (pA!= pB) {pA = pA == null? headB : pA.next;pB = pB == null? headA : pB.next;}return pA;}public static void main(String[] args) {// node listNode node8 = new Node(8);Node node4 = new Node(4);Node node5 = new Node(5);node8.next = node4;node4.next = node5;// A listNode headA = new Node(4);Node nodeA1 = new Node(1);headA.next = nodeA1;nodeA1.next = node8;// B ListNode headB = new Node(5);Node nodeB1 = new Node(6);Node nodeB2 = new Node(1);headB.next = nodeB1;nodeB1.next = nodeB2;nodeB2.next = node8;IntersectionNode intersectionNode = new IntersectionNode();Node startInter = intersectionNode.getStartInter(headA, headB);System.out.println(startInter);}
}

二、对称二叉树

Tree

package org.example.YangQianGuan;class Tree{int data;Tree left;Tree right;public Tree(int data, Tree left, Tree right) {this.data = data;this.left = left;this.right = right;}
}public class IsMirrorTree {public boolean isMirror(Tree root) {if (root == null) {return true;}return this.isMirrorCore(root.left, root.right);}public boolean isMirrorCore(Tree left, Tree right) {if (left == null && right == null) {return true;}if (left == null || right == null) {return false;}if (left.data != right.data) {return false;}return isMirrorCore(left.left, right.right) && isMirrorCore(left.right, right.left);}public static void main(String[] args) {Tree right4 = new Tree(3, null, null);Tree right3 = new Tree(4, null, null);Tree right2 = new Tree(2, right3, right4);Tree left4 = new Tree(4, null, null);Tree left3 = new Tree(3, null, null);Tree left2 = new Tree(2, left3, left4);Tree node1 = new Tree(1, left2, right2);IsMirrorTree isMirrorTree = new IsMirrorTree();boolean mirror = isMirrorTree.isMirror(node1);System.out.println(mirror);}}

三、轮转数组

package org.example.YangQianGuan;import java.util.Arrays;
import java.util.List;public class RightRunArr {public static void main(String[] args) {int[] nums = {1, 2, 3, 4, 5, 6, 7};int n = 3;RightRunArr rightRunArr = new RightRunArr();int[] core = rightRunArr.core(nums, n);System.out.println(Arrays.toString(core));}private int[] core(int[] nums, int n) {int[] res = new int[nums.length];int bit = n % nums.length;System.out.println(bit);int[] left = Arrays.copyOfRange(nums, 0, nums.length - bit);int[] right = Arrays.copyOfRange(nums, nums.length - bit, nums.length);int i = 0;for (int i1 : right) {res[i] = i1;i++;}for (int i1 : left) {res[i] = i1;i++;}return res;}
}

四、接雨水

题目描述

给定 n 个非负整数表示每个宽度为 1 的柱子的高度图，计算按此排列的柱子，下雨之后能接多少雨水。

示例 1：

输入：height = [0,1,0,2,1,0,1,3,2,1,2,1]
输出：6
解释：上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度图，在这种情况下，可以接 6 个单位的雨水（蓝色部分表示雨水）。 
示例 2：输入：height = [4,2,0,3,2,5]   输出：9

提示：

n == height.length
1 <= n <= 2 * 
0 <= height[i] <= 

package org.example.YangQianGuan;/*** 暴力求解法* 总水量 = 除了左右两个边界，【1，lenth-1】所有点位能接水的量的总和* 题目转化为如何求解每个点位的接水量 = Math.min(次点左边最大高度，此点右边最大高度）-heiht（i）*/
public class TrapSolution {public int trap(int[] height) {if (height.length<=1) {return 0;}int totleWater = 0;int leftMax = 0, rightMax = 0;for (int i = 1; i < height.length-1; i++) {// left maxfor (int j = 0; j < i; j++) {leftMax = Math.max(leftMax, height[j]);}// right maxfor (int j = i+1; j < height.length; j++) {rightMax = Math.max(rightMax, height[j]);}totleWater += Math.min(leftMax,rightMax)-height[i];}return totleWater;}public static void main(String[] args) {TrapSolution solution = new TrapSolution();// 测试用例 1int[] height1 = {0,1,0,2,1,0,1,3,2,1,2,1};System.out.println(solution.trap(height1)); // 输出: 6// 测试用例 2int[] height2 = {4,2,0,3,2,5};System.out.println(solution.trap(height2)); // 输出: 9}
}

五、字母异位词分组

给你一个字符串数组，请你将 字母异位词 组合在一起。可以按任意顺序返回结果列表。

字母异位词 是由重新排列源单词的所有字母得到的一个新单词。

示例 1:输入: strs = ["eat", "tea", "tan", "ate", "nat", "bat"] 输出: [["bat"],["nat","tan"],["ate","eat","tea"]]

示例 2:输入: strs = [""] 输出: [[""]]

示例 3:输入: strs = ["a"] 输出: [["a"]]

提示：

1 <= strs.length <= 
0 <= strs[i].length <= 100
strs[i] 仅包含小写字母

package org.example.YangQianGuan;import java.util.*;public class GroupAnagrams {public static void main(String[] args) {String[] s = {"eat", "tea", "tan", "ate", "nat", "bat"};GroupAnagrams groupAnagrams = new GroupAnagrams();String[] res = groupAnagrams.getGroup(s);}private String[] getGroup(String[] s) {int length = s.length;if (length==0) {return new String[0];}Map<String, List<String>> map = new HashMap<>();for (String string : s) {char[] charArray = string.toCharArray();Arrays.sort(charArray);String s1 = new String(charArray);List<String> orDefault = map.getOrDefault(s1, new ArrayList<>());orDefault.add(string);map.put(s1, orDefault);}String res[] = new String[map.size()];int n = 0;for (Map.Entry<String, List<String>> stringListEntry : map.entrySet()) {List<String> value = stringListEntry.getValue();String[] array = value.toArray(String[]::new);res[n] = Arrays.toString(array);n++;}System.out.println(Arrays.toString(res));return res;}
}