leecode更新
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dingjiawen
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7e0a828387
commit
5e970a95f0
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<groupId>org.example</groupId>
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<artifactId>Leecode</artifactId>
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<version>1.0-SNAPSHOT</version>
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<build>
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<plugins>
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<plugin>
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<groupId>org.apache.maven.plugins</groupId>
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<artifactId>maven-compiler-plugin</artifactId>
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<configuration>
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<source>8</source>
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<target>8</target>
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</configuration>
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</plugin>
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</plugins>
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</build>
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<dependencies>
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<!-- <dependency>-->
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<!-- <groupId>org.projectlombok</groupId>-->
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@ -0,0 +1,92 @@
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package com.markilue.leecode.hashtable;
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import org.junit.Test;
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import java.util.Arrays;
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import java.util.HashSet;
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import java.util.Iterator;
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import java.util.Set;
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/**
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* @BelongsProject: Leecode
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* @BelongsPackage: com.markilue.leecode.hashtable
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* @Author: dingjiawen
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* @CreateTime: 2022-09-07 11:47
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* @Description:
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* TODO 力扣349题:求两个数组的交集
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* 给定两个数组 nums1 和 nums2 ，返回 它们的交集 。输出结果中的每个元素一定是 唯一 的。我们可以 不考虑输出结果的顺序 。
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* @Version: 1.0
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*/
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public class Intersection {
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@Test
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public void test(){
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int[] nums1 = {1, 2, 2, 1};
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int[] nums2 = {2, 2};
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System.out.println(Arrays.toString(intersection(nums1, nums2)));
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}
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/**
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* 维护一个HashSet存储一个nums数组的数据,遍历另一个数组,如果有就保留
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* 速度击败99.13%，内存击败83.58%
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* @param nums1
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* @param nums2
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* @return
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*/
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public int[] intersection(int[] nums1, int[] nums2) {
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HashSet<Integer> integers = new HashSet<>();
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for (int i : nums1) {
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integers.add(i);
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}
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int[] result=new int[integers.size()];
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int count=0;
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for (int i : nums2) {
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if(integers.contains(i)){
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integers.remove(i);
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result[count++]=i;
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}
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}
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int[] result1=new int[count];
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for (int i = 0; i < result1.length; i++) {
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result1[i]=result[i];
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}
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return result1;
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}
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/**
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* 代码随想录方法.使用两个HashSet
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* 速度击败26%，内存击败68.81%
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* @param nums1
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* @param nums2
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* @return
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*/
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public int[] intersection1(int[] nums1, int[] nums2) {
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if (nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0) {
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return new int[0];
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}
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Set<Integer> set1 = new HashSet<Integer>();
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Set<Integer> resSet = new HashSet<Integer>();
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//遍历数组1
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for (int i : nums1) {
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set1.add(i);
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}
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//遍历数组2的过程中判断哈希表中是否存在该元素
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for (int i : nums2) {
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if (set1.contains(i)) {
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resSet.add(i);
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}
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}
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//将结果几何转为数组
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return resSet.stream().mapToInt(x -> x).toArray();
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}
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}
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File diff suppressed because one or more lines are too long
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@ -0,0 +1,123 @@
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package com.markilue.leecode.hashtable;
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import org.junit.Test;
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import java.lang.reflect.Array;
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import java.util.Arrays;
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import java.util.HashMap;
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/**
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* @BelongsProject: Leecode
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* @BelongsPackage: com.markilue.leecode.hashtable
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* @Author: dingjiawen
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* @CreateTime: 2022-09-07 12:45
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* @Description:
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* TODO leecode第1题:两数之和：
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* 给定一个整数数组 nums和一个整数目标值 target，请你在该数组中找出 和为目标值 target 的那两个整数，并返回它们的数组下标。
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* 你可以假设每种输入只会对应一个答案。但是，数组中同一个元素在答案里不能重复出现。
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* 你可以按任意顺序返回答案。
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* @Version: 1.0
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*/
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public class TwoSum {
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@Test
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public void test(){
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int[] nums = {2, 7, 11, 15};
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int target = 9;
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System.out.println(Arrays.toString(twoSum(nums, target)));
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}
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@Test
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public void test1(){
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int[] nums = {3,2,4};
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int target = 6;
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System.out.println(Arrays.toString(twoSum(nums, target)));
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}
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@Test
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public void test2(){
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int[] nums = {3,3};
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int target = 6;
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System.out.println(Arrays.toString(twoSum(nums, target)));
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}
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@Test
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public void test3(){
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int[] nums = {3,2,3};
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int target = 6;
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System.out.println(Arrays.toString(twoSum(nums, target)));
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}
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/**
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* 自己的思路:维护一个hashMap<数组位置,需要的值>
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* 速度击败53.1%，内存击败7.31%
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* @param nums
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* @param target
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* @return
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*/
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public int[] twoSum(int[] nums, int target) {
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HashMap<Integer, Integer> map = new HashMap<>();
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int[] result=new int[2];
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for (int i = 0; i < nums.length; i++) {
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map.put(target-nums[i],i);
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}
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int last=0;
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int index=0;
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boolean flag=false;
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for (int i = 0; i < nums.length; i++) {
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if(flag){
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if(nums[i]==last){
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result[0]=i;
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result[1]=index;
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return result;
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}
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}
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if (map.containsKey(nums[i])) {
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//如果进了这里还没有返回，那么可能是两数相等的情况
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flag=true;
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last=nums[i];
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index=i;
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if(map.get(nums[i])!=map.get(target-nums[i])){
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result[0]=map.get(nums[i]);
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result[1]=map.get(target-nums[i]);
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return result;
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}
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}
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}
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return result;
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}
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/**
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* 官方的思路:维护一个hashMap<数组位置,需要的值>，代码更加简洁,只使用一次循环,避免了很多的重复判断
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* 速度击败73.35%，内存击败54.57%
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* @param nums
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* @param target
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* @return
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*/
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public int[] twoSum1(int[] nums, int target) {
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HashMap<Integer, Integer> map = new HashMap<>();
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for (int i = 0; i < nums.length; i++) {
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if (map.containsKey(target-nums[i])) {
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return new int[]{i,map.get(target-nums[i])};
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}
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map.put(nums[i],i);
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}
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return new int[0];
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}
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}
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