遠景智能-秋招面試手撕代碼
1.尋找重複數
class Solution {
public int findDuplicate(int[] nums) {
Arrays.sort(nums);
for (int i = 1; i < nums.length; i++) {
if (nums[i] == nums[i-1]) {
return nums[i];
}
}
return -1;
}
}
時間複雜度：O(nlgn)
空間複雜度：O(1)(or O(n))，
class Solution {
public int findDuplicate(int[] nums) {
Set<Integer> seen = new HashSet<Integer>();
for (int num : nums) {
if (seen.contains(num)) {
return num;
}
seen.add(num);
}
return -1;
}
}
時間複雜度：O(n)
空間複雜度：O(n)
2.缺失數字
class Solution {
public int missingNumber(int[] nums) {
Arrays.sort(nums);
// 判斷 n 是否出現在末位
if (nums[nums.length-1] != nums.length) {
return nums.length;
}
// 判斷 0 是否出現在首位
else if (nums[0] != 0) {
return 0;
}
for(int i=0;i<nums.length;i++){
if(nums[i]-i==1){
return i;
}
}
return -1;
}
}
時間複雜度：O(nlogn)
空間複雜度：O(1)/O(n)
class Solution {
public int missingNumber(int[] nums) {
int missing = nums.length;
for (int i = 0; i < nums.length; i++) {
missing ^= i ^ nums[i];
}
return missing;
}
}
時間複雜度：O(n)
空間複雜度：O(1)
class Solution {
public int missingNumber(int[] nums) {
int expectedSum = nums.length*(nums.length + 1)/2;
int actualSum = 0;
for (int num : nums) actualSum += num;
return expectedSum - actualSum;
}
}
時間複雜度：O(n)
空間複雜度：O(1)
3.根據先序中序遍歷重建二叉樹
class Solution {
public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
if (pre ==null||in==null) {
return null;
}
if (pre.length == 0||in.length == 0) {
return null;
}
if (pre.length != in.length) {
return null;
}
TreeNode root= new TreeNode(pre[0]);
for (int i = 0; i < pre.length; i++) {
if (pre[0]==in[i]) {
root.left = reConstructBinaryTree(Arrays.copyOfRange(pre, 1, i+1), Arrays.copyOfRange(in, 0, i));
root.right = reConstructBinaryTree(Arrays.copyOfRange(pre, i+1, pre.length), Arrays.copyOfRange(in, i+1,in.length));
System.out.print(root.val+",");
}
}
return root;
}
}
4.兩個有序數組求中位數
class Solution {
public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
int n=nums1.length;
int m=nums2.length;
ArrayList<Integer> list=new ArrayList<Integer>();
for (int i = 0; i < n; i++) {
list.add(nums1[i]);
}
for (int i = 0; i < m; i++) {
list.add(nums2[i]);
}
Collections.sort(list);
int sum=list.size();
double res=0.0;
if (sum%2==0) {//偶數
int k=sum/2;
res=(double)(list.get(k)+list.get(k-1))/2;
}else {//奇數
res=list.get(sum/2);
}
return res;
}
}
5.兩個棧模擬隊列
public class MyQueue<E>{
private Stack<E> s1=new Stack<E>();
private Stack<E> s2=new Stack<E>();
public sychronized void put(E e){
s1.push(e);
}
public sychronized void pop(){
if(s2.isEmpty()){
while(!s1.isEmpty()){
s2.push(s1.pop());
}
}
return s2.pop();
}
public sychronized boolean empty(){
return s1.isEmpty()&&s2.isEmpty();
}
}
6.約瑟夫環
public static void yuesefu(int totalNum, int countNum) {
// 初始化人數
List<Integer> start = new ArrayList<Integer>();
for (int i = 1; i <= totalNum; i++) {
start.add(i);
}
//從第K個開始計數
int k = 0;
while (start.size() >0) {
k = k + countNum;
//第m人的索引位置
k = k % (start.size()) - 1;
// 判斷是否到隊尾
if (k < 0) {
System.out.println(start.get(start.size()-1));
start.remove(start.size() - 1);
k = 0;
} else {
System.out.println(start.get(k));
start.remove(k);
}
}
}
7.連續子數組的最大和
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int[] nums = new int[N];
for (int i = 0; i < N; i++) {
nums[i] = sc.nextInt();
}
int[] sums = new int[N];
sums[0] = nums[0];
int max = sums[0];
for (int i = 1; i < N; i++) {
sums[i] = Math.max(nums[i], sums[i - 1] + nums[i]);
max = Math.max(max, sums[i]);
}
System.out.println(max);
}
8.二叉樹的右視圖
public List<Integer> rightSideView(TreeNode root) {
if (root == null) return new ArrayList<>();
List<Integer> res = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
for (int i = 0; i<size; i++) {
TreeNode node = queue.poll();
if (i == (size-1)) {
res.add(node.val);
}
if (node.left != null) {
queue.add(node.left);
}
if (node.right != null) {
queue.add(node.right);
}
}
}
return res;
}
9.實心菱形
public static void print07() {
for(int x = -5;x<=5;x++) {
for(int y = -5;y<=5;y++) {
if((x>0?x:-x)+(y>0?y:-y)<=5) {
System.out.print("*");
}else {
System.out.print(" ");
}
}
System.out.println();
}
}
10.空心菱形
public static void main(String[] args) {
int s=4;
for(int i = 1;i<=s;i++) {//控制行數
for(int k = 1;k<=s-i;k++) {//空格的個數
System.out.print(" ");
}
for(int j = 1;j<=2*i-1;j++) {//控制星星個數的時候和行有關
if(j==1||j==2*i-1) {
System.out.print("*");
}else {
System.out.print(" ");
}
}
System.out.println();
}
for(int i = s-1;i>=1;i--) {//控制行數
for(int k = 1;k<=s-i;k++) {//空格的個數
System.out.print(" ");
}
for(int j = 1;j<=2*i-1;j++) {//控制星星個數的時候和行有關
if(j==1||j==2*i-1) {
System.out.print("*");
}else {
System.out.print(" ");
}
}
System.out.println();
}
}
11.遠景筆試第一題
public class Main {
public static List<Integer> findClosestElements(int[] arr, int k, int x) {
int size = arr.length;
int left = 0;
int right = size - 1;
int removeNums = size - k;
while (removeNums > 0) {
if (x - arr[left] <= arr[right] - x) {
right--;
} else {
left++;
}
removeNums--;
}
List<Integer> res = new ArrayList<>();
for (int i = left; i < left + k; i++) {
res.add(arr[i]);
}
return res;
}
public static void main(String[] args) {
int[] arr = {0, 0, 1, 2, 3, 3, 4, 7, 7, 8};
int k = 3;
int x = 5;
System.out.println(findClosestElements(arr, k, x));
}
}
12.非遞歸調用二叉樹
（1）前序遍歷
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res=new ArrayList<>();
Stack<TreeNode> stack=new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node=stack.pop();
if(node==null) continue;
res.add(node.val);
stack.push(node.right);//先右後左，保證左子樹先進行遍歷。
stack.push(node.left);
}
return res;
}
（2）後序遍歷
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> res=new ArrayList<>();
Stack<TreeNode> stack=new Stack<>();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node=stack.pop();
if(node==null) continue;
res.add(node.val);
stack.push(node.left);
stack.push(node.right);
}
Collections.reverse(res);
return res;
}
（3）中序遍歷
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res=new ArrayList<>();
Stack<TreeNode> stack=new Stack<>();
if(root==null)return res;
TreeNode cur=root;
while(cur!=null||!stack.isEmpty()){
while(cur!=null){
stack.push(cur);
cur=cur.left;
}
TreeNode node=stack.pop();
res.add(node.val);
cur=node.right;
}
return res;
}
13.層次遍歷二叉樹
public class Solution {
public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
ArrayList<Integer> list = new ArrayList<Integer>();
if(root==null) return list;
LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while(!queue.isEmpty()){
TreeNode node = queue.poll();
list.add(node.val);
if(node.left!=null){
queue.addLast(node.left);
}
if(node.right!=null){
queue.addLast(node.right);
}
}
return list ;
}
}
14.層次遍歷打印二叉樹，每層打印一行
public List<List<Integer>> levelOrder(TreeNode root) {
if(root == null)
return new ArrayList<>();
List<List<Integer>> res = new ArrayList<>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
while(!queue.isEmpty()){
int count = queue.size();
List<Integer> list = new ArrayList<Integer>();
while(count > 0){
TreeNode node = queue.poll();
list.add(node.val);
if(node.left != null)
queue.add(node.left);
if(node.right != null)
queue.add(node.right);
count--;
}
res.add(list);
}
return res;
}