1、IP转换成整数及整数转换成IP
Java实现：
package com.mian.demo;
public class IpToInt {
public static void main(String[] args) {
String ip="192.168.12.90";
int ipInt=ipToInt(ip);
System.out.println(ipInt);
String ipStr=intToIp(ipInt);
System.out.println(ipStr);
}
private static String intToIp(int ip){
StringBuilder sb=new StringBuilder();
for(int i=3;i>=0;--i){
int ipa=(ip>>(8*i))&(0xff);
System.out.print(ipa+" ");
sb.append(String.valueOf(ipa)+".");
}
return new String(sb).substring(0,sb.length()-1);
}
private static int ipToInt(String ip){
int res=0;
String[] arr=ip.split("\\.");
for(int i=0;i<arr.length;++i){
int ipa=Integer.parseInt(arr[i]);
System.out.print(ipa+" ");
res=(res<<8)|ipa;//用+也可以,但是位运算更快
}
return res;
}
}
C++实现：
1 #include <iostream>
2 #include <string>
3
4 using namespace std;
5
6 int ip_to_int(string ip)
7 {
8 size_t idx = 0;
9 string str = "";
10 int ret = 0;
11
12 while (true)
13 {
14 idx = ip.find_first_of('.');
15 str = ip.substr(0, idx);
16 int ipa = atoi(str.c_str());
17 ret = (ret << 8) | ipa;//用+也可以,但是位运算更快
18 ip = ip.substr(idx + 1);
19 if (idx == string::npos)
20 break;
21 }
22
23 return ret;
24 }
25
26 string int_to_ip(int ip)
27 {
28 string str = "";
29 for (int i = 3; i >= 0; --i)
30 {
31 int ipa = (ip >> (8 * i))&(0xff);
32 str.append(to_string(ipa) + ".");
33 }
34
35 return str.substr(0, str.size() - 1);
36 }
37
38 int main()
39 {
40 string ip = "192.168.12.90";
41 int p = ip_to_int(ip);
42 cout << p << endl;
43 cout << int_to_ip(p) << endl;
44
45 return 0;
46 }
2、给一个字符串表示IP地址，检测是否合法
Java实现：
package com.mian.demo;
import java.util.Scanner;
public class IsValidIp {
public static void main(String[] args){
Scanner sc=new Scanner(System.in);
while(sc.hasNext()) {
String input = sc.next();
System.out.println(isValidIp(input));
}
}
private static boolean isValidIp(String ip){
int size=ip.length();
if(ip.isEmpty()||size<7||size>15){
return false;
}
String[] arr=ip.split("\\.");
if(arr.length!=4){
return false;
}
for(int i=0;i<arr.length;++i){
if(arr[i].length()>1&&arr[i].charAt(0)=='0'){
return false;
}
for(int j=0;j<arr[i].length();++j){
if(arr[i].charAt(j)<'0'||arr[i].charAt(j)>'9'){
return false;
}
}
}
for(int i=0;i<arr.length;++i){
int tmp=Integer.parseInt(arr[i]);
if(i==0){
if(tmp<1||tmp>255){
return false;
}
}else{
if(tmp<0||tmp>255){
return false;
}
}
}
return true;
}
}
C++实现：
1 #include<iostream>
2 #include<vector>
3 #include<string>
4
5 using namespace std;
6
7 bool isValidIp(string &ip)
8 {
9 int size = ip.size();
10 if (ip.empty() || size < 7 || size>15)
11 return false;
12 vector<string> arr;
13 size_t idx = 0;
14 while (true)
15 {
16 idx = ip.find_first_of('.');
17 arr.push_back(ip.substr(0, idx));
18 ip = ip.substr(idx + 1);
19 if (idx == string::npos)
20 break;
21 }
22 if (arr.size() != 4)
23 return false;
24 for (int i = 0; i < arr.size(); ++i)
25 {
26 if (arr[i].size() > 1 && arr[i][0] == '0')
27 return false;
28 for (int j = 0; j < arr[i].size(); ++j)
29 if (arr[i][j]<'0' || arr[i][j]>'9')
30 return false;
31 }
32 for (int i = 0; i < arr.size(); ++i)
33 {
34 int tmp = atoi(arr[i].c_str());
35 if (i == 0)
36 {
37 if (tmp < 1 || tmp>255)
38 return false;
39 }
40 else
41 {
42 if (tmp < 0 || tmp>255)
43 return false;
44 }
45 }
46 return true;
47 }
48
49 int main()
50 {
51 string str;
52 while (cin >> str)
53 cout << isValidIp(str) << endl;
54
55 return 0;
56 }
3、杨辉三角
Java实现：
package com.mian.demo;
import java.util.ArrayList;
import java.util.List;
public class YangHuiSanJiao {
public static void main(String[] args) {
int[][] res=yangHuiSanJiao(10);
for(int i=0;i<res.length;++i){
for(int k=0;k<res.length-i;++k){
System.out.print(" ");
}
for(int j=0;j<res[i].length;++j){
System.out.print(res[i][j]+" ");
}
System.out.println();
}
}
private static int[][] yangHuiSanJiao(int n){
int[][] res=new int[n][n];
for(int i=0;i<n;++i){
res[i]=new int[i+1];
}
for(int i=0;i<n;++i){
res[i][0]=1;
res[i][i]=1;
for(int j=1;j<i;++j){
res[i][j]=res[i-1][j]+res[i-1][j-1];
}
}
return res;
}
}
C++实现：
1 #include<iostream>
2 #include<vector>
3
4 using namespace std;
5
6 vector<vector<int>> yanghui(int n)
7 {
8 vector<vector<int>> res(n, vector<int>());
9 for (int i = 0; i < n; ++i)
10 {
11 res[i].resize(i + 1);
12 res[i][0] = 1;
13 res[i][i] = 1;
14 }
15 for (int i = 1; i < n; ++i)
16 for (int j = 1; j < i; ++j)
17 res[i][j] = res[i - 1][j - 1] + res[i - 1][j];
18
19 return res;
20 }
21
22 int main()
23 {
24 int n;
25 while (cin >> n)
26 {
27 vector<vector<int>> res = yanghui(n);
28 for (int i = 0; i < n; ++i)
29 {
30 for (int k = 0; k < n - i; ++k)
31 cout << " ";
32 for (int j = 0; j < res[i].size(); ++j)
33 if (j == 0)
34 cout << res[i][j];
35 else
36 cout << " " << res[i][j];
37 cout << endl;
38 }
39 }
40
41 return 0;
42 }
4、定义一个函数求字符串长度，要求该函数体中不能声明任何一个变量（不使用额外变量，实现strlen函数）
1 #include <iostream>
2
3 using namespace std;
4
5 int getLen(char *str)
6 {
7 if (*str == '\0')
8 return 0;
9
10 return getLen(str + 1) + 1;
11 }
12
13 int getLength(char *str)
14 {
15 return *str ? (getLength(str + 1) + 1) : 0;
16 }
17
18
19 int main()
20 {
21 char str[] = "abcde";
22 int len1 = getLen(str);
23 cout << len1 << endl;
24 int len2 = getLength(str);
25 cout << len2 << endl;
26
27 return 0;
28 }
5、两个整数序列，第一个序列表示栈的压入顺序，请判断第二个序列是否为该栈的弹出顺序。
Java实现：
package com.mian.demo;
import java.util.Stack;
public class IsPopOrder {
public static void main(String[] args) {
int[] pushV={ 1,2,3,4,5 };
int[] popV1={ 4,5,3,2,1 };
int[] popV2={ 4,3,5,1,2 };
boolean t=isPopOrder(pushV,popV1);
boolean f=isPopOrder(pushV,popV2);
System.out.println(t+" "+f);
}
private static boolean isPopOrder(int[] pushV,int[] popV){
int psize=pushV.length;
int osize=popV.length;
if(psize!=osize){
return false;
}
Stack<Integer> stk=new Stack<>();
int popIndex=0;
for(int i=0;i<psize;++i){
stk.push(pushV[i]);
while(!stk.isEmpty()&&stk.peek()==popV[popIndex]){
++popIndex;
stk.pop();
}
}
return stk.isEmpty();
}
}
C++实现：
1 #include<iostream>
2 #include<vector>
3 #include<stack>
4
5 using namespace std;
6
7 bool isPopOrder(vector<int> &pushV, vector<int> &popV)
8 {
9 int psize = pushV.size();
10 int osize = popV.size();
11 if (psize == 0 || osize == 0 || psize != osize)
12 return false;
13 stack<int> stk;
14 int popIndex = 0;
15 for (int i = 0; i < psize; ++i)
16 {
17 stk.push(pushV[i]);
18 while (!stk.empty() &&stk.top() == popV[popIndex])
19 {
20 ++popIndex;
21 stk.pop();
22 }
23 }
24 return stk.empty();
25 }
26
27 int main()
28 {
29 vector<int> pushV = { 1,2,3,4,5 };
30 vector<int> popV1 = { 4,5,3,2,1 };
31 vector<int> popV2 = { 4,3,5,1,2 };
32 bool t = isPopOrder(pushV, popV1);
33 bool f = isPopOrder(pushV, popV2);
34 cout << t << endl;
35 cout << f << endl;
36
37 return 0;
38 }
6、逆序对数
Java实现：
package com.mian.demo;
public class InversePairs {
static int pairNum;
public static void main(String[] args){
int[] arr={ 7,5,6,4 };
InversePairs inversePairs=new InversePairs();
int res=inversePairs.inversePairs(arr);
System.out.println(res);
}
private int inversePairs(int[] arr){
pairNum=0;
if(arr!=null){
mergeSort(arr,0,arr.length-1);
}
return pairNum;
}
private void mergeSort(int[] arr,int low,int high){
int mid=(low+high)>>1;
if(low<high){
mergeSort(arr,low,mid);
mergeSort(arr,mid+1,high);
mergeCore(arr,low,mid,high);
}
}
private void mergeCore(int[] arr,int left,int mid,int right){
int[] tmp=new int[right-left+1];
int i=left;
int j=mid+1;
int k=0;
while(i<=mid&&j<=right){
if(arr[i]<arr[j]) {
tmp[k++] = arr[i++];
}else{
tmp[k++]=arr[j++];
pairNum+=mid-i+1;
}
}
while(i<=mid){
tmp[k++]=arr[i++];
}
while(j<=right){
tmp[k++]=arr[j++];
}
for(int m=0;m<tmp.length;++m){
arr[left+m]=tmp[m];
}
}
}
C++实现：
View Code
7、找零问题
Java实现：
package com.mian.demo;
import java.lang.reflect.Field;
public class FindMinMoney {
public static void main(String[] args) {
FindMinMoney fm=new FindMinMoney();
fm.findMinMoney(18,new int[]{1,2,5,8,10});
}
private void findMinMoney(int money,int[] coin){
int[] coinNum=new int[money+1];
int[] coinValue=new int[money+1];
coinNum[0]=0;
for(int i=1;i<=money;++i){
int minNum=i;
int usedMoney=0;
for(int j=0;j<coin.length;++j){
if(i>=coin[j]){
if(coinNum[i-coin[j]]+1<=minNum&&(i==coin[j]||coinValue[i-coin[j]]!=0)){
minNum=coinNum[i-coin[j]]+1;
usedMoney=coin[j];
}
}
}
coinNum[i]=minNum;
coinValue[i]=usedMoney;
}
if(coinValue[money]==0){
System.out.println("找不开零钱");
}else{
System.out.println("需要最少硬币个数为："+coinNum[money]);
System.out.print("硬币分别为: ");
while(money>0){
System.out.print(coinValue[money]+" ");
money-=coinValue[money];
}
}
}
}
C++实现：
1 #include<iostream>
2
3 using namespace std;
4
5 //money需要找零的钱,coin可用的硬币,n硬币种类
6 void findMin(int money, int *coin, int n)
7 {
8 int *coinNum = new int[money + 1]();//存储1...money找零最少需要的硬币的个数
9 int *coinValue = new int[money + 1]();//最后加入的硬币，方便后面输出是哪几个硬币
10 coinNum[0] = 0;
11
12 for (int i = 1; i <= money; i++)
13 {
14 int minNum = i;//i面值钱，需要最少硬币个数
15 int usedMoney = 0;//这次找零，在原来的基础上需要的硬币
16 for (int j = 0; j<n; j++)
17 {
18 if (i >= coin[j])//找零的钱大于这个硬币的面值
19 {
20 if (coinNum[i - coin[j]] + 1 <= minNum && (i == coin[j] || coinValue[i - coin[j]] != 0))
21 {
22 minNum = coinNum[i - coin[j]] + 1;
23 usedMoney = coin[j];
24 }
25 }
26 }
27 coinNum[i] = minNum;
28 coinValue[i] = usedMoney;
29 }
30
31 //输出结果
32 if (coinValue[money] == 0)
33 cout << "找不开零钱" << endl;
34 else
35 {
36 cout << "需要最少硬币个数为：" << coinNum[money] << endl;
37 cout << "硬币分别为:";
38 while (money>0)
39 {
40 cout << coinValue[money] << " ";
41 money -= coinValue[money];
42 }
43 }
44 delete[]coinNum;
45 delete[]coinValue;
46 }
47 int main()
48 {
49 int Money = 18;
50 int coin[] = { 1,2,5,9,10 };
51 findMin(Money, coin, 5);
52
53 return 0;
54 }
8、字符串翻转
Java实现：
package com.mian.demo;
public class ReverseString {
public static void main(String[] args) {
String str="abcdef";
ReverseString rs=new ReverseString();
String res=rs.reverseString(str);
System.out.println(res);
}
private String reverseString(String str){
int n=str.length();
if(n==0||str.isEmpty()){
return str;
}
char[] chars=str.toCharArray();
for(int i=0,j=n-1;i<j;++i,--j){
char c=chars[i];
chars[i]=chars[j];
chars[j]=c;
}
return new String(chars);
}
}
C++实现：
1 #include<iostream>
2
3 void reverseStr(char *str, int n)
4 {
5 for (int i = 0, j = n - 1; i < j; ++i, --j)
6 {
7 char c = str[i];
8 str[i] = str[j];
9 str[j] = c;
10 }
11 }
12
13 int main()
14 {
15 char c[] = "abcdef";
16 int size = strlen(c);
17 std::cout << c << std::endl;
18 reverseStr(c, size);
19 std::cout << c << std::endl;
20
21 return 0;
22 }
9、判断两个日期之间相隔的天数
Java实现：
package com.mian.demo;
public class DiffDay {
public static void main(String[] args) {
int year1 = 2017, month1 = 9, day1 = 23;
int year2 = 2018, month2 = 3, day2 = 25;
DiffDay diffDay=new DiffDay();
int diff=diffDay.getDiffDay(year1,month1,day1,year2,month2,day2);
System.out.println(diff);
}
private int getDiffDay(int year1,int month1,int day1,int year2,int month2,int day2){
int d1=getDay(year1,month1,day1);
int d2=getDay(year2,month2,day2);
int res=0;
if(year1==year2){
res=Math.abs(d1-d2);
}else if(year1>year2){
for(int i=year2+1;i<year1;++i){
if(isLeap(i)){
res+=366;
}else{
res+=365;
}
}
res+=d1;
if(isLeap(year2)){
res+=(366-d2);
}else{
res+=(365-d2);
}
}else{
for(int i=year1+1;i<year2;++i){
if(isLeap(i)){
res+=366;
}else{
res+=365;
}
}
res+=d2;
if(isLeap(year1)){
res+=(366-d1);
}else{
res+=(365-d1);
}
}
return res;
}
private int getDay(int year,int month,int day){
int days=0;
int d=0;
for(int i=1;i<month;++i){
switch (i){
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:
d=31;
days+=d;
break;
case 2:
if(isLeap(year)){
d=29;
}else{
d=28;
}
days+=d;
break;
case 4:
case 6:
case 9:
case 11:
d=30;
days+=d;
break;
}
}
return days+day;
}
private boolean isLeap(int year){
if((year%4==0&&year%100==0)||year%400==0){
return true;
}
return false;
}
}
C++实现：
1 #include<iostream>
2 #include<iostream>
3
4 using namespace std;
5
6 bool isLeap(int year)
7 {
8 if ((year % 4 == 0 && year % 100)|| year % 400==0)
9 return true;
10 return false;
11 }
12
13 int getDay(int year, int month, int day)
14 {
15 int res = 0;
16 int d = 0;
17 for (int i = 1; i < month; ++i)
18 {
19 switch (i)
20 {
21 case 1:
22 case 3:
23 case 5:
24 case 7:
25 case 8:
26 case 10:
27 case 12:
28 d = 31;
29 res += d;
30 break;
31 case 2:
32 if (isLeap(year))
33 d = 29;
34 else
35 d = 28;
36 res += d;
37 break;
38 case 4:
39 case 6:
40 case 9:
41 case 11:
42 d = 30;
43 res += d;
44 break;
45 }
46 }
47 return res + day;
48 }
49
50 int main()
51 {
52 int year1 = 2017, month1 = 9, day1 = 23;
53 int year2 = 2018, month2 = 3, day2 = 25;
54 int d1 = getDay(year1, month1, day1);
55 int d2 = getDay(year2, month2, day2);
56 int res = 0;
57 if (year1 == year2)
58 cout << std::abs(d1-d2) << endl;
59 else
60 {
61 if (year1 > year2)
62 {
63 for (int i = year2 + 1; i < year1; ++i)
64 {
65 if (isLeap(i))
66 res += 366;
67 else
68 res += 365;
69 }
70 res += d1;
71 if (isLeap(year2))
72 res += (366 - d2);
73 else
74 res += (365 - d2);
75 }
76 else if (year2 > year1)
77 {
78 for (int i = year1 + 1; i < year2; ++i)
79 {
80 if (isLeap(i))
81 res += 366;
82 else
83 res += 365;
84 }
85 res += d2;
86 if (isLeap(year1))
87 res += (366 - d1);
88 else
89 res += (365 - d1);
90 }
91 cout << res << endl;
92 }
93
94 return 0;
95 }
10、给定A, B两个整数，不使用除法和取模运算，求A/B的商和余数。
Java实现：
package com.mian.demo;
public class Divide {
public static void main(String[] args){
divide(10,3);
}
private static void divide(int a,int b){
for(int i=2;i<a;++i){
if(i*b>a){
System.out.println((i-1)+" "+(a-((i-1)*b)));
break;
}
}
}
}
C++实现：
1 #include<iostream>
2
3 using namespace std;
4
5 void divide(int a, int b)
6 {
7 for (int i = 2; i < a; ++i)
8 if (i*b > a)
9 {
10 cout << (i - 1) << " " << (a - (i - 1)*b) << endl;
11 break;
12 }
13 }
14
15 int main()
16 {
17 divide(10, 3);
18
19 return 0;
20 }
11、压缩驼峰字符串
Java实现:
package com.mian.demo;
public class CamelAndSnakeString {
public static void main(String[] args){
String s = "get_set_where";
System.out.println(camelString(s));
String str="getSetWhere";
System.out.println(snakeString(str));
}
private static String camelString(String str){
int size=str.length();
if(size==0||str.isEmpty()){
return str;
}
StringBuilder sb=new StringBuilder();
for(int i=0;i<size;++i){
if(str.charAt(i)=='_'){
continue;
}
if(i!=0&&str.charAt(i-1)=='_'){
sb.append(new Character(str.charAt(i)).toString().toUpperCase());
}else {
sb.append(str.charAt(i));
}
}
return new String(sb);
}
private static String snakeString(String str){
int size=str.length();
if(size==0||str.isEmpty()){
return str;
}
StringBuilder sb=new StringBuilder();
for(int i=0;i<size;++i){
if(str.charAt(i)>='A'&&str.charAt(i)<='Z'){
sb.append('_'+new Character(str.charAt(i)).toString().toLowerCase());
}else{
sb.append(str.charAt(i));
}
}
return new String(sb);
}
}
C++实现：
1 #include<iostream>
2 #include<string>
3
4 using namespace std;
5
6 string snakeString(string &str)
7 {
8 int size = str.size();
9 if (size == 0 || str.empty())
10 return "";
11 string res = "";
12 for (int i = 0; i < str.size(); ++i)
13 {
14 if (islower(str[i]))
15 res += str[i];
16 else
17 {
18 res += '_';
19 res += tolower(str[i]);
20 }
21 }
22 return res;
23 }
24
25 string camelString(string &s)
26 {
27 int size = s.size();
28 if (size == 0 || s.empty())
29 return "";
30 string res = "";
31 for (int i = 0; i < s.size(); ++i)
32 {
33 if (s[i]=='_')
34 continue;
35 if (i!=0&&s[i - 1] == '_')
36 res += toupper(s[i]);
37 else
38 res += s[i];
39 }
40 return res;
41 }
42
43 int main()
44 {
45 string str = "getSetWhere";
46 cout << snakeString(str) << endl;
47 string s = "get_set_where";
48 cout << camelString(s) << endl;
49
50 return 0;
51 }
12、除去s1中与s2中相同的字符
Java实现：
package com.mian.demo;
import java.util.Arrays;
public class Squeeze {
public static void main(String[] args) {
String str1 = "abcde";
String str2 = "ace";
System.out.println(squeeze(str1, str2));
}
private static String squeeze(String str1, String str2) {
int i = 0, j = 0, k = 0;
StringBuilder sb = null;
while (j < str2.length()) {
sb = new StringBuilder();
while (i < str1.length()) {
if (str1.charAt(i) != str2.charAt(j)) {
sb.append(str1.charAt(i));
}
++i;
}
str1 = new String(sb);
i = k = 0;
++j;
}
return str1;
}
}
C++实现：
1 #include <stdio.h>
2
3 void squeeze(char s1[], char s2[])
4 {
5 int i, j, k;
6 i = j = k = 0;
7
8 while (s2[j] != '\0')
9 {
10 while (s1[i] != '\0')
11 {
12 if (s2[j] != s1[i])
13 s1[k++] = s1[i];
14 i++;
15 }
16 s1[k] = '\0';
17 i = k = 0;
18 j++;
19 }
20 }
21
22 int main()
23 {
24 char s1[10] = { 'a','b','c','d','e',0 };
25 char s2[10] = { 'a','c','e',0 };
26
27 squeeze(s1, s2);
28 printf("%s\n", s1);
29 return 0;
30 }
13、给定n个数，要求比较1.5n次同时找出最大值和最小值
分析：要求比较次数为1.5n，使用一般的逐个遍历每个元素然后判断其是否为最大最小值是需要2n次比较的。现在考虑采用，每次从数组中取出两个元素，判断其大小，然后再分别判断其是否是最大或最小值，这样一次处理两个元素，每一次比较3次，最终的比较次数就是n/2*3=1.5n。
Java实现：
package com.mian.demo;
public class FindMaxMin {
public static void main(String[] args){
int num[]={2, 4, 5, 6, 8, 3, 7, 1, 9, 10 };
findMaxMin(num);
}
private static void findMaxMin(int[] arr){
int size=arr.length;
if(size==0||arr==null){
return;
}
int max=Integer.MIN_VALUE;
int min=Integer.MAX_VALUE;
int i=0;
int j=size-1;
int tmin,tmax;
int count=0;
while(i<j){
if(arr[i]<arr[j]){
tmax=arr[j];
tmin=arr[i];
++count;
}else{
tmax=arr[i];
tmin=arr[j];
++count;
}
if(min>tmin){
min=tmin;
}
if(max<tmax){
max=tmax;
}
count+=2;
++i;
--j;
}
System.out.println("The max number is "+max);
System.out.println("The min number is "+min);
System.out.println("Compare number is "+count);
}
}
C++实现：
1 #include<cstdio>
2 #include<iostream>
3
4 void find_max_min(int num[], int n)
5 {
6 int i = 0, j = n - 1;
7 int max = INT_MIN;
8 int min = INT_MAX;
9 int tmax, tmin;
10 int count = 0; /*用来统计比较次数*/
11
12 while (i < j)
13 {
14 if (num[i] < num[j])
15 {
16 tmax = num[j];
17 tmin = num[i];
18 ++count;
19 }
20 else
21 {
22 tmax = num[i];
23 tmin = num[j];
24 ++count;
25 }
26 if (min > tmin)
27 min = tmin;
28 if (max < tmax)
29 max = tmax;
30 count += 2; /*上面的两次比较*/
31 ++i;
32 --j;
33 }
34 printf("The max number is: %d.\n", max);
35 printf("The min number is: %d.\n", min);
36 printf("Compare number is: %d.\n", count);
37 }
38
39 int main()
40 {
41 int num[10] = { 2, 4, 5, 6, 8, 3, 7, 1, 9, 10 };
42 find_max_min(num, 10);
43
44 return 0;
45 }
14、按照map的value进行排序，并按value有序输出
package com.mian.demo;
import java.util.*;
public class MapValueSort {
public static void main(String[] args){
Map<String,Integer> map=new HashMap<>();
map.put("three",3);
map.put("nine",9);
map.put("zero",0);
map.put("five",5);
map.put("two",2);
map.put("six",6);
MapValueSort m=new MapValueSort();
Map<String,Integer> res=m.mapValueSort(map);
for(String key:res.keySet()){
System.out.println(key+" "+res.get(key));
}
}
private <T extends Comparable> Map<String,T> mapValueSort(Map<String,T> map){
int size=map.size();
if(size==0||map==null){
return map;
}
List<Map.Entry<String,T>> list=new ArrayList<>();
for(Map.Entry<String,T> entry:map.entrySet()){
list.add(entry);
}
Collections.sort(list, new Comparator<Map.Entry<String, T>>() {
@Override
public int compare(Map.Entry<String, T> o1, Map.Entry<String, T> o2) {
return o1.getValue().compareTo(o2.getValue());
}
});
LinkedHashMap<String,T> m=new LinkedHashMap<>();
// Map<String,T> m=new HashMap<>();
// Map<String,T> m=new TreeMap<>();
for(Map.Entry<String,T> entry:list){
m.put(entry.getKey(),entry.getValue());
}
return m;
}
}
15、找出数组里出现次数超过总数1/3的数(可能存在一个也可能存在两个)
class Solution {
public List<Integer> majorityElement(int[] nums) {
List<Integer> res=new ArrayList<Integer>();
if(nums==null||nums.length==0){
return res;
}
int a=nums[0],cnta=0;
int b=nums[0],cntb=0;
for (int num : nums) {
if (num == a) {
++cnta;
}else if(num == b) {
++cntb;
}else if(cnta == 0) {
a = num;
cnta=1;
}else if(cntb == 0) {
b = num;
cntb=1;
}else{
--cnta;
--cntb;
}
}
cnta=0;
cntb=0;
for(int num:nums){
if(num==a){
++cnta;
}else if(num==b){
++cntb;
}
}
if(cnta>nums.length/3){
res.add(a);
}
if(cntb>nums.length/3){
res.add(b);
}
return res;
}
}
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