剑指offer~编程小哥令狐
一、数组类~
03、数组中重复的数字
class Solution
{
public void swap(int[] nums,int i,int j)
{
int temp=nums[i];
nums[i]=nums[j];
nums[j]=temp;
}
public int findRepeatNumber(int[] nums)
{
int n=nums.length;
for(int num:nums)
if(num<0||num>=n)
return -1;
for(int i=0;i<n;i++)
{
while(nums[i]!=i&&nums[i]!=nums[nums[i]])
swap(nums,i,nums[i]);
if(nums[i]!=i&&nums[i]==nums[nums[i]])
return nums[i];
}
return -1;
}
}
04、二维数组中的查找
- 暴力查找
class Solution {
public boolean findNumberIn2DArray(int[][] matrix, int target) {
int i,j;
if((matrix==null||matrix.length==0)||(matrix.length==1&&matrix[0].length==0))
return false;
for (i=0;i<matrix.length;i++)
for (j=0;j<matrix[0].length;j++)
if(matrix[i][j]==target)
return true;
return false;
}
}
- 线性查找
class Solution {
public boolean findNumberIn2DArray(int[][] matrix, int target) {
if((matrix==null||matrix.length==0)||(matrix.length==1&&matrix[0].length==0))
return false;
int i=0,j=matrix[0].length-1;
while(i<=matrix.length-1&&j>=0){
if(target==matrix[i][j]) return true;
if(target>matrix[i][j]) i++;
else if(target<matrix[i][j]) j--;
}
return false;
}
}
05、顺时针打印矩阵
class Solution {
public int[] spiralOrder(int[][] matrix) {
//判断边界
if(matrix.length==0)
return new int[0];//因为返回值是数组，所以实例化一个数组进行返回
//右下左上
int left=0,right=matrix[0].length-1,up=0,down=matrix.length-1;
int num=0;
int []res=new int[(right+1)*(down+1)];
while(true){
//右
for(int i=left;i<=right;i++)
res[num++]=matrix[up][i];
if(++up>down) break;///判断边界
//下
for(int i=up;i<=down;i++)
res[num++]=matrix[i][right];
if(--right<left) break;
//左
for(int i=right;i>=left;i--)
res[num++]=matrix[down][i];
if(--down<up) break;
//上
for(int i=down;i>=up;i--)
res[num++]=matrix[i][left];
if(++left>right) break;
}
return res;
}
}
06、在排序数组中查找数字
- 暴力查找法
class Solution {
public int search(int[] nums, int target) {
int res=0;
for(int i=0;i<nums.length;i++)
if(nums[i]==target)
res++;
return res;
}
}
- 二分查找法【有序数组优先考虑二分查找】
class Solution {
public int search(int[] nums, int target) {
int left = 0, right = nums.length;
int ans = 0;
while (left < right){
int mid = left + ((right - left)/2);
if (nums[mid] >= target){
right = mid;
} else {
left = mid + 1;
}
}
while (left < nums.length && nums[left++] == target){
ans++;
}
return ans;
}
}
07、剑指 Offer 53 - II. 0～n-1中缺失的数字
class Solution {
public int missingNumber(int[] nums) {
int left = 0, right = nums.length;
while(left < right){
int mid = left + (right - left) / 2;
if (nums[mid] == mid) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
}
二、链表类~
06、从尾到头打印链表
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public int[] reversePrint(ListNode head) {
ListNode temp=head;
int size=0;
while(temp!=null){
size++;
temp=temp.next;
}
int[] res=new int[size];
temp=head;
for(int i=size-1;i>=0;i--){
res[i]=temp.val;
temp=temp.next;
}
return res;
}
}
07、删除链表节点
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode deleteNode(ListNode head, int val) {
if(head.val==val) return head.next;
ListNode pre=head;
ListNode cur=head.next;
while(cur.val!=val&&cur!=null){
pre=cur;
cur=cur.next;
}
if(cur.val==val){
pre.next=cur.next;
}
return head;
}
}
08、删除倒数第k个节点
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode getKthFromEnd(ListNode head, int k) {
ListNode current=head;
int n=0;
while(current!=null){
n++;
current=current.next;
}
if(k>n) return null;
int aim=n-k+1;
int ptr=1;
current=head;
while(ptr<aim){
current=current.next;
ptr++;
}
return current;
}
}
09、反转链表
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode reverseList(ListNode head) {
ListNode pre=null;
ListNode cur=head;
while (cur!=null){
ListNode temp=cur.next;
cur.next=pre;
pre=cur;
cur=temp;
}
return pre;
}
}
10、复杂链表的复制
/*
// Definition for a Node.
class Node {
int val;
Node next;
Node random;
public Node(int val) {
this.val = val;
this.next = null;
this.random = null;
}
}
*/
class Solution {
public Node copyRandomList(Node head) {
Node cur=head;
HashMap<Node,Node> map=new HashMap<>();
while(cur!=null){
map.put(cur,new Node(cur.val));
cur=cur.next;
}
cur=head;
while(cur!=null){
map.get(cur).next=map.get(cur.next);
map.get(cur).random=map.get(cur.random);
cur=cur.next;
}
return map.get(head);
}
}
11、两个链表的第一个公共点
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
ListNode a=headA;
ListNode b=headB;
while(a!=b){
if(a==null)
a=headB;
else
a=a.next;
if(b==null)
b=headA;
else
b=b.next;
}
return a;
}
}
三、哈希表~
12、数组中重复的数字
class Solution
{
public int findRepeatNumber(int[] nums)
{
Set<Integer> set=new HashSet<Integer>();
int repeat=0;
for(int i=0;i<nums.length;i++){
if(!set.add(nums[i])){
repeat=nums[i];
break;
}
}
return repeat;
}
}
13、最长不含重复字符的子字符串
class Solution {
public int lengthOfLongestSubstring(String s) {
HashMap<Character,Integer> hash=new HashMap<>();
int res=0,left=0;
for (int i=0;i<s.length();i++){
char c=s.charAt(i);
//判断c是否出现过
//abcabcbb
if (hash.containsKey(c)){
left=Math.max(left,hash.get(c)+1);
}
hash.put(c,i);
res=Math.max(res,i-left+1);
}
return res;
}
}
14、第一个只出现一次的字符
class Solution {
public char firstUniqChar(String s) {
HashMap<Character,Boolean> hash=new HashMap<>();
char []ch=s.toCharArray();
for(char c:ch)
hash.put(c,!hash.containsKey(c));
for(char c:ch)
if(hash.get(c))
return c;
return ' ';
}
}
四、栈~
15、用两个栈实现队列
import java.util.HashMap;
import java.util.Stack;
class CQueue {
//栈 先进后出
//队列 先进先出
//Stack<Integer> stk1,stk2;
Stack<Integer> stk1=new Stack<Integer>();
Stack<Integer> stk2=new Stack<Integer>();
int size;
public CQueue() {
//stk1=new Stack<Integer>();
//stk2=new Stack<Integer>();
size=0;
}
public void appendTail(int value) {
//插入一个元素
//stk1保存 底部存新插入的 顶部存老的
while(!stk1.isEmpty())
stk2.push(stk1.pop());
stk1.push(value);
while (!stk2.isEmpty())
stk1.push(stk2.pop());
size++;
}
public int deleteHead() {
//删除队列首部元素
//删除栈顶
if(size==0)
return -1;
int res=stk1.pop();
size--;
return res;
}
}
/**
* Your CQueue object will be instantiated and called as such:
* CQueue obj = new CQueue();
* obj.appendTail(value);
* int param_2 = obj.deleteHead();
*/
