LeetCode刷题day001 (Jieky)

LeetCode第一题

/*
Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1]
*/

import java.util.Arrays;
import java.util.HashMap; 
import java.util.Map; 

public class TwoSum{
	
	// 暴力破解法，寻找两个数的和等于目标数
	public static int[] towSum01(int[] nums,int target){
		int ans[] = new int[2];
		for (int i=0;i<nums.length;i++){
			for (int j=i+1;j<nums.length;j++){
				if (nums[i]+nums[j] == target){
					ans[0] = i;
					ans[1] = j;
					return ans;
				}
			}
		}
		return ans;
	}
	
	// 先将数组放入HashMap，只需判断sub在不在hash的key里就可以了，这种解法利用了hash数据结构的快速访问的特性
	public static int[] towSum02(int[] nums,int target){
		int ans[] = new int[2];
		Map<Integer,Integer> map = new HashMap<>();
		for (int i=0;i<nums.length;i++){
			map.put(nums[i],i);
		}
		
		for (int i=0;i<nums.length;i++){
			int sub = target - nums[i];
			if (map.containsKey(sub) && map.get(sub) != i){
				ans[0] = i;
				ans[1] = map.get(sub);
				return ans;
			}
		}
		return ans;
	}
	
	// 这种实现的优化在于，寻找的另一个数一定不包含在map里
	public static int[] towSum03(int[] nums,int target){
		int ans[] = new int[2];
		Map<Integer,Integer> map = new HashMap<>();
		for (int i=0;i<nums.length;i++){
			map.put(nums[i],i);
			int sub = target - nums[i];
			if (map.containsKey(sub)){
				ans[0] = i;
				ans[1] = map.get(sub);
			}
		}
		return ans;
	}
	
	public static void main(String[] args){
		int nums[] = {2,7,11,15};
		int target = 9;
		int result[] = towSum03(nums,target);
		Arrays.sort(result);
		System.out.println(Arrays.toString(result));
	}
}

LeetCode第二题

/*
You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
*/

import java.util.ArrayList;
import java.util.Collections;

class ListNode{
	int val;
	ListNode next;
	ListNode(int x){val=x;}
}
	
public class AddTwoNumbers{
	
	public static void main(String[] args){
		// 生成第一个链表
		ListNode L1_head = new ListNode(4);
		ListNode temp01 = L1_head;
		temp01.next = new ListNode(4);
		temp01 = temp01.next;
		temp01.next = new ListNode(3);
		temp01 = temp01.next;
		temp01.next=null;
		
		// 生成第二个链表
		ListNode L2_head = new ListNode(5);
		ListNode temp = L2_head;
		temp.next = new ListNode(6);
		temp = temp.next;
		temp.next = new ListNode(4);
		temp = temp.next;
		temp.next=null;
		
		// new一个实例对象
		AddTwoNumbers addTwoNumbers = new AddTwoNumbers();
		
		// 对输入链表进行逆序
		L1_head = addTwoNumbers.reverseListRecursion(L1_head);
		L2_head = addTwoNumbers.reverseListRecursion(L2_head);
		
		// 对输入链表再次进行逆序
		L1_head = addTwoNumbers.reverseList(L1_head);
		L2_head = addTwoNumbers.reverseList(L2_head);
		
		ListNode result = addTwoNumbers.addTwoNumbers(L1_head,L2_head);
		
		ArrayList<Integer> ids = new ArrayList<Integer>();
		while(result != null){
			ids.add(result.val);
			result = result.next;
		}
		//Collections.reverse(ids);
		System.out.println(ids);
	}
	
	//递归思想，可以从最后还剩两个节点的链表为例进行理解
	public ListNode reverseListRecursion(ListNode head){
		ListNode newHead;
		if(head==null||head.next==null ){
			return head;
		}
		//head.next 指向当前需逆序子链表的头节点，指向逆序后的尾节点
		newHead=reverseListRecursion(head.next); 
		//head.next 代表指向新链表的尾，将它的 next 置为 head，就是将 head 加到最后作为真正的尾
		head.next.next=head; 
		//之前的head现在变成了逆序链表真正的尾，其next指向null
		head.next=null;
		
		return newHead;
	}

	//迭代思想
	public ListNode reverseList(ListNode head){
		if (head==null){return null;}
		ListNode pre = null;
		ListNode next = null;
		while(head!=null){
			next = head.next;
			head.next = pre;
			pre = head;
			head = next;
		}
		return pre;
	}
	
	//两个链表的相加
	public ListNode addTwoNumbers(ListNode L1,ListNode L2){
		// 保存结果链表的头，头不赋值，以免后期做判断
		ListNode dummyHead = new ListNode(0);
		ListNode p = L1, q = L2, curr = dummyHead;
		int carray = 0;
		while(p!=null || q!=null){
			int x = (p!=null) ? p.val:0;
			int y = (q!=null) ? q.val:0;
			// 逐次相加并加上相应的进位
			int sum = x + y + carray;
			// 进位最大为1 9+9=18
			carray = sum/10;
			curr.next = new ListNode(sum%10);
			curr = curr.next;
			if (p!=null){p = p.next;}
			if (q!=null){q = q.next;}
		}
		
		//判断最后一位数，是否发生进位
		if (carray > 0){
			curr.next = new ListNode(carray);
			// 始终保证curr指向最后一个节点
			curr = curr.next;
		}
		// 最后一个节点的next指向null
		curr.next = null;
		// 之所以next，是因为链表头不存数据
		return dummyHead.next;
	}
}

本文地址：https://blog.csdn.net/qq_24964575/article/details/110941756