给定目标总和，找出给定数组中是否存在一对总和的元素
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import java.util.HashMap;
public class target
{
public static void hash(int []a,int sum)
{
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int i;
for (i = 0; i < a.length; ++i)
map.put(a[i], sum-a[i]);
for (i = 0; i < a.length; ++i)
if(map.containsValue(a[i]) && map.get(a[i])!=null)
{
System.out.println("("+a[i]+","+map.get(a[i])+")");
map.remove(a[i]);
}
}
public static void main(String[] args)
{
int []a={1, 2, 13, 34, 9, 3, 23, 45, 8, 7, 8, 3, 2};
hash(a,11);
}
}
我想知道上面的解决方案是否有更好，更有效的解决方案。这种复杂性是n。我可以做得更好吗？
答案
您的实现错过了重复的对。
你可以
- 对数组进行排序
- 从一开始就为每个元素迭代 计算所需的补数（sum - element） 进行反向二进制搜索（从排序数组的末尾）查找该精确值 如果找到，请删除两者
它归结为观察到，元素排序：
n1 < n2 < n3 < n4 < n5 < n6
最可能的对从两端到中间对称。现在，最坏的情况仍然很糟糕，但至少你没有哈希表开销
另一答案
正如我评论的那样，你的sollution不是O（N），因为containsValue搜索存储在HashMap中的所有值。为了解决这个问题，我使用了您的解决方案：
public static void newVersion(int[] a, int sum){
HashMap<Integer, Boolean> map = new HashMap<Integer, Boolean>();
for (int i= 0; i< a.length; i++) {
map.put(sum - a[i], true);
}
for (int i = 0; i < a.length; i++) {
if (map.containsKey(a[i]) && map.get(a[i])) {
System.out.println("("+(sum-a[i])+","+a[i]+")");
map.put(a[i], false);
map.put(sum-a[i], false);
}
}
}
在第一步，它存储每个整数的“补码值”，在第二步，它检查补码是否存在。如果存在，请将两个对标记为已使用。
这种复杂性是：
* O(N) for the first looping
* O(N) * (O(1) + O(1)) for the second loop and the containsValue and get.
* Finally: O(N) + O(N) .:. O(N) solution,
另一答案
我有这个问题的以下解决方案。时间复杂度应该是O（N），因为HashMap操作put，get和keySet是O（1）。
import java.util.HashMap;
import java.util.Map;
/**
* Find a pair of numbers in an array given a target sum
*
*
*/
public class FindNums {
public static void findSumsForTarget(int[] input, int target)
{
// just print it instead of returning
Map<Integer, String> myMap = populateMap(input);
// iterate over key set
for (Integer currKey : myMap.keySet()) {
// find the diff
Integer diff = target - currKey;
// check if diff exists in the map
String diffMapValue = myMap.get(diff);
if(diffMapValue!=null)
{
// sum exists
String output = "Sum of parts for target " + target + " are " + currKey + " and " + diff;
System.out.println(output);
return; // exit; we're done - unless we wanted all the possible pairs and permutations
}
// else
// keep looking
}
System.out.println("No matches found!");
}
private static Map<Integer, String> populateMap(int[] input)
{
Map<Integer,String> myMap = new HashMap<Integer,String>();
for (int i = 0; i < input.length; i++) {
String currInputVal = myMap.get(input[i]);
if(currInputVal!=null) // value already exists
{
// append current index location to value
currInputVal = currInputVal + ", " + i;
// do a put with the updated value
myMap.put(input[i], currInputVal);
}
else
{
myMap.put(input[i], Integer.toString(i)); // first argument is autoboxed to Integer class
}
}
return myMap;
}
// test it out!
public static void main(String[] args)
{
int[] input1 = {2,3,8,12,1,4,7,3,8,22};
int[] input2 = {1,2,3,4,5,6,7,8,9,10};
int[] input3 = {2,-3,8,12,1,4,7,3,8,22};
int target1 = 19;
int target2 = 16;
// test
FindNums.findSumsForTarget(input1, target1);
FindNums.findSumsForTarget(input1, -1);
FindNums.findSumsForTarget(input2, target2);
FindNums.findSumsForTarget(input3, target1);
}
}
另一答案
import java.util.*;
import java.io.*;
class hashsum
{
public static void main(String arg[])throws IOException
{
HashMap h1=new HashMap();
h1.put("1st",new Integer(10));
h1.put("2nd",new Integer(24));
h1.put("3rd",new Integer(12));
h1.put("4th",new Integer(9));
h1.put("5th",new Integer(43));
h1.put("6th",new Integer(13));
h1.put("7th",new Integer(5));
h1.put("8th",new Integer(32));
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter no.");
int no=Integer.parseInt(br.readLine());
Iterator i=h1.entrySet().iterator();
boolean flag=false;
while(i.hasNext())
{
Map.Entry e1=(Map.Entry)i.next();
Integer n1=(Integer)e1.getValue();
Iterator j=h1.entrySet().iterator();
while(j.hasNext())
{
Map.Entry e2=(Map.Entry)j.next();
Integer n2=(Integer)e2.getValue();
if(no==(n1+n2))
{
System.out.println("Pair of elements:"+n1 +" "+n2);
flag=true;
}
}
}
if(flag==false)
System.out.println("No pairs");
}
}
另一答案
public static void hash1(int[] a, int num) {
Arrays.sort(a);
// printArray(a);
int top = 0;
int bott = a.length - 1;
while (top < bott) {
while (a[bott] > num)
bott--;
int sum = a[top] + a[bott];
if (sum == num) {
System.out.println("Pair " + a[top] + " " + a[bott]);
top++;
bott--;
}
if (sum < num)
top++;
if (sum > num)
bott--;
}
}
另一答案
解决方案：O（n）时间和O（log（n））空间。
public static boolean array_find(Integer[] a, int X)
{
boolean[] b = new boolean[X];
int i;
for (i=0;i<a.length;i++){
int temp = X-a[i];
if(temp >= 0 && temp < X) //make sure you are in the bound or b
b[temp]=true;
}
for (i=0;i<a.length;i++)
if(a[i]<X && b[a[i]]) return true;
return false;
}
另一答案
递归地找到其总和是来自给定数组的目标和的子集。
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class Main {
public static Set<List<Integer>> set = new HashSet<>();
public static void main(String[] args) {
int[] biggerArray = {1, 2, 1, 1};
int targetedSum = 3;
findSubset(biggerArray, targetedSum);
}
public static void findSubset(int[] biggerArray, int targetedSum) {
for (int i = 0; i < biggerArray.length; i++) {
List<Integer> subset = new ArrayList<>();
if (biggerArray[i] > targetedSum)
continue;
else
subset.add(biggerArray[i]);
if (i + 1 < biggerArray.length)
find(subset, i, biggerArray, targetedSum, i);
}
System.out.println(set);
}
public static List<Integer> find(List<Integer> subset, int startIndex, final int[] biggerArray, final int targetedSum, final int skipIndex) {
if (skipIndex == startIndex) {
find(subset, startIndex + 1, biggerArray, targetedSum, skipIndex);
return null;
}
int subsetSum = findSumOfList(subset);
int remainedSum = targetedSum - subsetSum;
int i = startIndex;
if (remainedSum == 0) {
set.add(subset);
return null;
}
if ((startIndex < biggerArray.length) && (biggerArray[startIndex] == remainedSum)) {
List<Integer> temp = new ArrayList<Integer>(subset);
temp.add(biggerArray[i]);
set.add(temp);
}
else if ((startIndex < biggerArray.length) && (biggerArray[startIndex] < remainedSum)) {
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