【算法】数据结构面试算法题目
1 数组去重
python实现
#调用内置函数去重
def func(str):
len1=len(str)
len2=len(list(set(str)))
print("去重后的结果是：",list(set(str)),"\t去重个数是：",(len1-len2))
#for 循环去重
def func1(str):
nums=[]
for n in str:
if n not in nums:
nums.append(n)
print("去重后的结果是：",sorted(nums),"\t去重个数是：",(len(str)-len(nums)))
# while去重
def func2(str):
len1=len(str)
for n in str:
while str.count(n)>1:
del str[str.index(n)]
print("去重后的结果是：",sorted(str),"\t去重个数是：",(len1-len(str)))
str=[1,3,2,4,2,4,1,6,4,5]
func2(str)
字典去重
#调用内置函数去重
def func(str):
len1=len(str.values())
len2=len(list(set(str.values())))
print("去重后的结果是：",list(set(str.values())),"\t去重个数是：",(len1-len2))
#for 循环去重
def func1(str):
nums=[]
for n in str.values():
if n not in nums:
nums.append(n)
print("去重后的结果是：",sorted(nums),"\t去重个数是：",(len(str.values())-len(nums)))
str=[1,3,2,4,2,4,1,6,4,5]
dirc={1:1,2:2,3:3,4:3}
func2(dirc)
python字符串追加去重排序
#调用内置函数去重
def func(str1,str2):
print("去重后的结果是：",sorted(set(str1+str2)),"\t去重个数是：",(len(str1+str2)-len(set(str1+str2))))
#for 循环去重
def func1(str1,str2):
nums=[]
str=str1+str2
for n in str:
if n not in nums:
nums.append(n)
print("去重后的结果是：",sorted(nums),"\t去重个数是：",(len(str)-len(nums)))
# while去重
def func2(str1,str2):
str=str1+str2
for n in str:
while str.count(n)>1:
del str[str.index(n)]
print("去重后的结果是：",sorted(str),"\t去重个数是：",(len(str1+str2)-len(str)))
str1=['很好','不错','很好','Very','Book','I','Love','I']
str2=['Java','C#','Python','C#']
func2(str1,str2)
Java实现
import java.util.LinkedList;
import java.util.List;
public class func {
public static void unique(String[] str){
// array_unique
List<String> list = new LinkedList<String>();
for(int i = 0; i < str.length; i++) {
if(!list.contains(str[i])) {
list.add(String.valueOf(str[i]));
}
}
System.out.print("去重后的结果："+list+"\t 共去重个数："+(str.length-list.size()));
}
public static void main(String[] args){
String[] str={"1","3","2","4","2","4","1","6","4","5"};
unique(str);
}
}
set实现
import java.util.HashSet;
import java.util.Set;
public class func {
public static void unique(String[] str){
Set<String> set = new HashSet<String>();
for (int i=0; i<str.length; i++) {
set.add(str[i]);
}
System.out.print("去重后的结果："+set+"\t 共去重个数："+(str.length-set.size()));
}
public static void main(String[] args){
String[] str={"Python","SQL","C#","Java","C","Python","R","Matlab","C++","SQL"};
unique(str);
}
}
java字符串追加去重实现
/* package whatever; // don't place package name! */
import java.util.*;
import java.util.Set;
import java.util.HashSet;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
/**
* 两个数组合并去重
* 2016年11月5日12:43:34
* 白宁超
* str1 long[] 数组1
* str2 long[] 数组2
*/
public static void disstr(long[] str1,long[] str2 ){
if(str1.length<=0||str2.length<=0){
return;
}
long[] str= new long[str1.length+str2.length];
System.arraycopy(str1, 0, str, 0, str1.length);
System.arraycopy(str2, 0, str, str1.length, str2.length);
Set<Long> set=new HashSet<Long>();
for(int i=0;i<str.length;i++){
set.add(str[i]);
}
System.out.print("追加去重后的数据是：\t"+set+"\n\t\t去重个数：\t"+(str.length-set.size()));
}
public static void main (String[] args) throws java.lang.Exception
{
long[] str1={10,20,30,50,10,60,40,20};
long[] str2={50,60,90,80,70};
disstr(str1,str2);
}
}
2 求数组中逆序对的总数，如输入数组1,2,3,4,5，6，7,0 逆序对7
Python实现
# 求数组中逆序对的总数，如输入数组1,2,3,4,5，6，7,0 逆序对7 def index(str3): res=0 if(len(str3)<0): return 0 if(len(str3)==0): return str3[-1] else: for i in str3: if(str3[i]>str3[-1]): res+=1 print(res) str3=[1,2,3,4,5,6,7,0] #func2(str1,str2) index(str3)
Java实现
/* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
public static void func2(){
int res=0;
int[] str={1,2,3,4,5,6,7,0};
for(int i=0;i<str.length;i++){
for(int j=i;j<str.length;j++){
if(str[i]>str[j]){
res++;
}
}
}
System.out.print(res);
}
public static void main (String[] args) throws java.lang.Exception
{
func2();
}
}
3 无序数组A，找到第K个最大值，复杂度小于O（NlgN）
Python实现
方法一：时间复杂度set()*sorted（）的复杂度
# 无序数组A，找到第K个最大值，复杂度小于O（NlgN）
def Maxnum(str,n):
if n>len(str) or n<0:
print("输入不合法")
return 0
numArr=sorted(set(str))
m=-n
print(numArr[m])
str3=[1,2,3,4,5,6,7,0]
Maxnum(str3,30)
方法二：时间复杂度O(n)+sorted（）复杂度
def Maxnum1(str,n):
newstr=[]
if n>len(str) or n<=0:
print("输入不合法")
return 0
else:
for i in str:
if i not in newstr:
newstr.append(i)
print(sorted(newstr)[-n])
str3=[1,2,3,4,5,6,7,0]
Maxnum1(str3,1)
方法三：时间复杂度O(nlogn)
def bnc_quick(arr,low,high): if low < high: key=arr[low] left=low right=high while low < high: while low < high and arr[high] >= key: high -= 1 arr[low]=arr[high] while low < high and arr[low] <= key: low += 1 arr[high]=arr[low] arr[low]=key bnc_quick(arr,left,low-1) bnc_quick(arr,low+1,right) arr=[20,10,30,40,100,60,90,210] print(arr) bnc_quick(arr,0,len(arr)-1) n=3 print(arr[-n])
方法四：时间复杂度O（n）
from random import randint def findKthMax(l,k): if k>len(l): return key=randint(0,len(l)-1) keyv=l[key] sl=[i for i in l[:key]+l[key+1:] if i<keyv] bl=[i for i in l[:key]+l[key+1:] if i>=keyv] if len(bl)==k-1: return keyv elif len(bl)>=k: return findKthMax(bl,k) else: return findKthMax(sl,k-len(bl)-1)
方法五：时间复杂度O(n)
def base_quick(arr,low,high): if low < high: key=arr[low] left=low right=high while low < high: while low < high and arr[high] >= key: high -= 1 arr[low]=arr[high] while low < high and arr[low] <= key: low += 1 arr[high]=arr[low] arr[low]=key return low def findmax(arr,k): length = len(arr) low = 0 high = length - 1 midkey = base_quick(arr, low, high) while midkey != k: if midkey > k: midkey = base_quick(arr, low, midkey - 1) elif midkey < k: midkey = base_quick(arr, midkey + 1, high) return arr[-k:]
Java实现
方法一：时间复杂度O(NlgN)
/* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
public static void quickSort(int[] arr,int low,int hight){
if(low < hight){
int key=arr[low];
int left=low;
int right=hight;
while(low < hight){
while(low < hight&&arr[hight] >= key){
hight--;
}
arr[low]=arr[hight];
while(low <hight&&arr[low] <= key){
low++;
}
arr[hight]=arr[low];
}
arr[low]=key;
quickSort(arr,left,low-1);
quickSort(arr,low+1,right);
}
}
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
int[] arr={1,3,4,6,7,99,45,26};
for(int i:arr){
System.out.print(i+" ");
}
System.out.println("\n**************************************");
quickSort(arr,0,arr.length-1);
for(int i:arr){
System.out.print(i+" ");
}
System.out.println("\n**************************************");
int k=3;
for(int i=0;i<arr.length;i++){
if(i==arr.length-k){
System.out.print("\n第"+k+"个最大数是："+arr[arr.length-k]);
}
System.out.println("\n*************** }***********************");
int n=3;
System.out.print("前"+n+"个最大数是：\n");
for(int i=arr.length-1;i>=arr.length-n;i--){
System.out.print(arr[i]+" ");
}
}
}
4 排序算法
Python实现冒泡排序
def buttle(arr): length = len(arr) for i in range(0,length): for j in range(i+1,length): if(arr[i]>arr[j]): arr[i],arr[j]=arr[j],arr[i] print(arr) str3=[1,2,13,4,15,6,7,0] buttle(str3)
Java实现冒泡排序
public static void buttle(int[] arr){
int temp=0;
for(int i=0;i<arr.length;i++){
for(int j=i+1;j<arr.length;j++){
if(arr[i]>arr[j]){
temp=arr[j];
arr[j]=arr[i];
arr[i]=temp;
}
}
}
for(int n:arr){
System.out.print(n+" ");
}
}
Python实现快排：
def bnc_quick(arr,low,high): if low < high: key=arr[low] left=low right=high while low < high: while low < high and arr[high] >= key: high -= 1 arr[low]=arr[high] while low < high and arr[low] <= key: low += 1 arr[high]=arr[low] arr[low]=key bnc_quick(arr,left,low-1) bnc_quick(arr,low+1,right) arr=[20,10,30,40,100,60,90,210] print(arr) bnc_quick(arr,0,len(arr)-1) print(arr)
Java实现快排：
/* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
public static void quickSort(int[] arr,int low,int hight){
if(low < hight){
int key=arr[low];
int left=low;
int right=hight;
while(low < hight){
while(low < hight&&arr[hight] >= key){
hight--;
}
arr[low]=arr[hight];
while(low <hight&&arr[low] <= key){
low++;
}
arr[hight]=arr[low];
}
arr[low]=key;
quickSort(arr,left,low-1);
quickSort(arr,low+1,right);
}
}
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
int[] arr={1,3,4,6,7,99,45,26};
for(int i:arr){
System.out.print(i+" ");
}
System.out.println("\n**************************************");
quickSort(arr,0,arr.length-1);
for(int i:arr){
System.out.print(i+" ");
}
}
}
简单选择排序：
def select(arr): for i in range(0,len(arr)): min=i for j in range(i+1,len(arr)): if arr[min] > arr[j]: min=j arr[min],arr[i]=arr[i],arr[min] return arr
5 不借助中间变量的两数字交换：
方法一：
private static void swapBySelf(int a, int b) {
// 在不引入其它变量的情况下交换两个数，利用两数之和来做
a = a+b; //a保存两数之和
b = a-b; //两数之和-b，即为a
a = a-b; //两数之和-b，此时的b已经变成了a，所以相当于sum-a=b
}
方法二：
//还有另一种方法,利用两数之差，即两数之间的距离
a = b-a; //a=两者的差
b = b-a; //b = 原来的b-两数的距离==原来的a
a = a+b; //最终的a=两者之差+原来的a==原来的b
System.out.println("swapBySelf second function:a="+a+",b="+b);//又换回来了
方法三：
//已知x^k^k==x，即一个数与任意一个数作两次异或运算都会变成原来的自己
private static void swapByXOR(int a, int b) {
// 在不引入其它变量的情况下交换两个数，利用异或来做
a = a^b; //a保存两数异或的中间结果
b = a^b; //a两次异或b就变成原来的a，并将其赋值给了b
a = a^b; //b两次异或a就变成原来的b，并且将其赋值给了a
System.out.println("swapByXOR first function:a="+a+",b="+b);
}