数据结构

数组

• 一组相同数据类型的集合

特点

1. 数组在内存中是连续分配的
2. 创建时要指明数组的大小
3. 数组名代表首地址,索引从0开始,到数组的长度-1
4. 数组一旦创建好,大小不可以改变
5. 使用索引
   1. 获取索引位置的值 arr[index]
   2. 修改 arr[index] = val
   3. 删除 (假删除)
   4. 遍历,将数组中的元素,依次打印出来

使用Java实现更高级的数组

package Arrays;import java.util.Random;public class MyArr<T> {private int capacity = 0;private int size = 0;private T[] arr;public MyArr(int capacity) {if (capacity < 0) this.capacity = 10; //if no right input, we will initial capacity 10this.capacity = capacity;this.arr = (T[]) new Object[capacity];}public int getCapacity() {return capacity;}public int getSize() {return size;}public T[] setCapacity(int capacity) {if (capacity < 0) {throw new RuntimeException("扩大小异常");}this.capacity = capacity;T[] newNum = (T[]) new Object[capacity];for (int i = 0; i < this.size; ++i) {newNum[i] = this.arr[i];}return newNum;}//增加元素public void add(T val) {if (this.size >= this.capacity) {this.arr = setCapacity(2 * this.capacity);}this.arr[this.size++] = val;}//删除元素public boolean removeByIndex(int index) {if (index < 0 || index > this.capacity) {throw new RuntimeException("数组越界");}for (int i = index; i < size - 1; ++i) {arr[i] = arr[i + 1];}size--;if (size < this.capacity / 4 && this.capacity > 4) {arr = setCapacity(this.capacity / 4);}return true;}//修改位置元素public void modify(int index, T val) {if (index < 0 || index > size - 1) {throw new RuntimeException("数组越界");}arr[index] = val;}//获取某元素位置public int locateVal(T val) {for (int i = 0; i < size; ++i) {if (arr[i] == val) {return i;//return index}}// if no find return -1return -1;}//打印元素@Overridepublic String toString() {StringBuffer stringBuffer = new StringBuffer();stringBuffer.append('[');for (int i = 0; i < this.size - 1; ++i) {stringBuffer.append(arr[i] + ",");}if(size>0) stringBuffer.append(arr[size - 1]);stringBuffer.append(']');return stringBuffer.toString();}}

对应习题

26. 删除有序数组中的重复项

class Solution {public int removeDuplicates(int[] nums) {/*    TreeSet<Integer> set = new TreeSet<>();for(int i=0;i<nums.length;i++){set.add(nums[i]);}Iterator<Integer> iterator = set.iterator();int temp=0;while(iterator.hasNext()){nums[temp] = iterator.next();temp++;}return set.size();*///使用双指针,优化/*int p=0;int q = 1;if(nums==null||nums.length==0)return 0;while(q<nums.length){if(nums[p]!=nums[q]){nums[p+1]=nums[q];p++;}q++;}return p+1;*///当数组根本不存在重复元素时,则上面的方法每次依然会进行重复的复制,显然这是没有必要的//再次优化int p=0;int q = 1;if(nums==null||nums.length==0)return 0;while(q<nums.length){if(nums[p]!=nums[q]){if(q-p>1)nums[p+1]=nums[q];//判断p++;}q++;}return p+1;}
}

1. 两数之和

class Solution {public int[] twoSum(int[] nums, int target) {int[] numsSum = new int[2];HashMap<Integer,Integer> hashMap = new HashMap<>();for(int i=0;i<nums.length;++i){if(hashMap.containsKey(target - nums[i])){return new int[]{i, hashMap.get(target - nums[i])};}hashMap.put(nums[i],i);}return null;// int[] numsSum = new int[2];// if(nums==null||nums.length<2)return numsSum;// int len = nums.length;// HashMap<Integer,Integer> hashMap = new HashMap<>();// for(int i=0;i<nums.length;++i){//     hashMap.put(nums[i],i);// }// int temp=0;// for(int i=0;i<nums.length;++i){//     temp = target - nums[i];//     if(hashMap.containsKey(temp)&&hashMap.get(temp)!=i){//         numsSum = new int[]{i, hashMap.get(temp)};//     }// }// return numsSum;}    }

class Solution:def twoSum(self, nums: List[int], target: int) -> List[int]:lens = len(nums)dic_num = {}for i in range(0,lens):if(target-nums[i]) in dic_num:return [i,dic_num.get(target-nums[i])]dic_num[nums[i]] = ireturn []