本文介绍了HashMap为所有Keys存储相同的值的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我想为唯一坐标存储不同的值。我使用整数数组将这些值存储在HashMap中的相应坐标，但每个键映射到上次计算的值。
代码：
import java.util。*;
import java.awt.Point;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;
public class Solution
{
@SuppressWarnings（empty-statement）
public static void main（String [] args）
{
扫描仪输入=新扫描仪（System.in）;
int n = in.nextInt（）;
int m = in.nextInt（）;
Integer [] ar = new Integer [3];
Map< Point，Integer []> map = new HashMap<>（）;
for（int a0 = 0; a0
Point p = new Point（in.nextInt（），in.nextInt（））;
int a = in.nextInt（）;
int b = in.nextInt（）; （map.containsKey（p））{
if（map.get（p）[2]<（a - b））{
ar [0] = a;
ar [1] = b;
ar [2] = a - b;
map.put（p，ar）;
}
} else {
ar [0] = a;
ar [1] = b;
ar [2] = a - b;
map.put（p，ar）;
}
}
Set< Entry< Point，Integer []>> set = map.entrySet（）;
List< Entry< Point，Integer []>> list = new ArrayList<>（set）; （Map.Entry< Point，Integer []>条目：list）
$ b System.out.println（entry.getKey（）+==== + Arrays.toString（entry.getValue（）））;
$ b}
}
输入：
0 1 1 1
1 2 2 4
2 0 1 2
结果：
java.awt.Point [x = 0，y = 1] ==== [1，2，-1]
java.awt.Point [x = 1，y = 2] ==== [1，2，-1]
java.awt.Point [x = 2，y = 0] ==== [1，2，-1]
解决方案
您只有 ar 数组的单个实例，您只需将其存储在多个键下。您需要每次创建一个新的数组实例：
import java.awt.Point;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
import java.util.Set;
public class Solution {
public static void main（String [] args）{
try（Scanner in = new Scanner（System.in））{
int n = in.nextInt（）;
int m = in.nextInt（）;
Map< Point，Integer []> map = new HashMap<>（）;
for（int a0 = 0; a0
Point p = new Point（in.nextInt（），in.nextInt（））;
int a = in.nextInt（）;
int b = in.nextInt（）;
//每个键的新数组
Integer [] ar = new Integer [3]; （map.containsKey（p））{
if（map.get（p）[2]<（a - b））{
ar [0] = a;
ar [1] = b;
ar [2] = a - b;
map.put（p，ar）;
}
} else {
ar [0] = a;
ar [1] = b;
ar [2] = a - b;
map.put（p，ar）;
}
}
Set< Entry< Point，Integer []>> set = map.entrySet（）;
List< Entry< Point，Integer []>> list = new ArrayList<>（set）; （Map.Entry< Point，Integer []>条目：列表）
b
System.out.println（entry.getKey（）+====+ Arrays。的toString（entry.getValue（）））;
$ b code
$ b 输出：
java.awt.Point [x = 0，y = 1] ==== [1，1，0]
java.awt.Point [x = 1，y = 2] ==== [2，4，-2]
java.awt.Point [x = 2，y = 0] === = [1，2，-1]
（还要注意 n 实际上并没有在任何地方被读取。）
为了更好地理解这个简化的例子：
import java.util.HashMap;
import java.util.Map;
public class SimpleSolution {
public static void main（String [] args）{
Map< Integer，Integer []> map = new HashMap<>（）;
Integer [] arr = new Integer [1];
arr [0] = 1;
map.put（1，arr）;
System.out.println（map）;
System.out.println（map.get（1）[0]）;
arr [0] = 2;
map.put（2，arr）;
System.out.println（map）;
System.out.println（map.get（1）[0]）;
$ / code $ / pre
$ b $输出：
{1 = [Ljava.lang.Integer; @ 2a139a55}
1
{1 = [Ljava.lang.Integer; @ 2a139a55， 2 = [Ljava.lang.Integer; @ 2a139a55}
2
您可以看到这里是
- 键1和2都具有相同的数组引用
- 设置
arr [0] 设置为2，键1的值在地图中也发生了变化
I want to store different values for unique coordinates.I am using integer array to store those values in HashMap to corresponding coordinates but every key maps to last calculated value.
Code :
import java.util.*;
import java.awt.Point;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Set;
public class Solution
{
@SuppressWarnings("empty-statement")
public static void main(String[] args)
{
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int m = in.nextInt();
Integer[] ar = new Integer[3];
Map<Point, Integer[]> map = new HashMap<>();
for (int a0 = 0; a0 < m; a0++) {
Point p = new Point(in.nextInt(), in.nextInt());
int a = in.nextInt();
int b = in.nextInt();
if (map.containsKey(p)) {
if (map.get(p)[2] < (a - b)) {
ar[0] = a;
ar[1] = b;
ar[2] = a - b;
map.put(p, ar);
}
} else {
ar[0] = a;
ar[1] = b;
ar[2] = a - b;
map.put(p, ar);
}
}
Set<Entry<Point, Integer[]>> set = map.entrySet();
List<Entry<Point, Integer[]>> list = new ArrayList<>(set);
for (Map.Entry<Point, Integer[]> entry : list)
System.out.println(entry.getKey() + " ==== " + Arrays.toString(entry.getValue()));
}
}
Input :
3 3
0 1 1 1
1 2 2 4
2 0 1 2
Result :
java.awt.Point[x=0,y=1] ==== [1, 2, -1]
java.awt.Point[x=1,y=2] ==== [1, 2, -1]
java.awt.Point[x=2,y=0] ==== [1, 2, -1]
解决方案 You only have one single instance of your ar array and you are just storing it under multiple keys. You need to create a new array instance each time:
import java.awt.Point;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.Scanner;
import java.util.Set;
public class Solution {
public static void main(String[] args) {
try (Scanner in = new Scanner(System.in)) {
int n = in.nextInt();
int m = in.nextInt();
Map<Point, Integer[]> map = new HashMap<>();
for (int a0 = 0; a0 < m; a0++) {
Point p = new Point(in.nextInt(), in.nextInt());
int a = in.nextInt();
int b = in.nextInt();
// New array for each key
Integer[] ar = new Integer[3];
if (map.containsKey(p)) {
if (map.get(p)[2] < (a - b)) {
ar[0] = a;
ar[1] = b;
ar[2] = a - b;
map.put(p, ar);
}
} else {
ar[0] = a;
ar[1] = b;
ar[2] = a - b;
map.put(p, ar);
}
}
Set<Entry<Point, Integer[]>> set = map.entrySet();
List<Entry<Point, Integer[]>> list = new ArrayList<>(set);
for (Map.Entry<Point, Integer[]> entry : list)
System.out.println(entry.getKey() + " ==== " + Arrays.toString(entry.getValue()));
}
}
}
Output:
java.awt.Point[x=0,y=1] ==== [1, 1, 0]
java.awt.Point[x=1,y=2] ==== [2, 4, -2]
java.awt.Point[x=2,y=0] ==== [1, 2, -1]
(Also note that n isn't actually read anywhere.)
For a better understanding consider this simplified example:
import java.util.HashMap;
import java.util.Map;
public class SimpleSolution {
public static void main(String[] args) {
Map<Integer, Integer[]> map = new HashMap<>();
Integer[] arr = new Integer[1];
arr[0] = 1;
map.put(1, arr);
System.out.println(map);
System.out.println(map.get(1)[0]);
arr[0] = 2;
map.put(2, arr);
System.out.println(map);
System.out.println(map.get(1)[0]);
}
}
Output:
{1=[Ljava.lang.Integer;@2a139a55}
1
{1=[Ljava.lang.Integer;@2a139a55, 2=[Ljava.lang.Integer;@2a139a55}
2
What you can see here is
- Key 1 and 2 both have the same array reference
- After setting
arr[0] to 2, the value for key 1 has also changed in the map
这篇关于HashMap为所有Keys存储相同的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！