blocks|key|4109501|text|public+static+int[]+twoSum(int[]+nums,+int+target)+{
++++++++int[]+sumnums+=+nums.clone();
++++++++//int[2]+sol+=+{0,1};
++++++++int[]+sol+=+new+int[]{0,1};

++++++++sol+=+new+int[2];
++++++++int+j=sumnums.length;+//+Every+recursion+j+will+be+initialized+as+sumnums.length+instead+of+J-1
++++++++int+t=target;

++++++++for(int+i=0;i<sumnums.length;i%2B%2B){
+++++++++++//+if+((sumnums[i]%2Bsumnums[j])==t){++
+++++++++++//+Remember+that+Java+arrays+start+at+index+0,+so+this+value(sumnums[j])+is+not+exist+in+the+array.
++++++++++++++if+((sumnums[i]%2Bsumnums[j-1])==t){
++++++++++++++++sol[0]=i;
++++++++++++++++sol[1]=j;
++++++++++++++++return+sol;
++++++++++++}
++++++++}
++++++++j=j-1;
++++++++twoSum(sumnums,t);
++++++++return+sol;

++++}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4109502|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>public static int[] twoSum(int[] nums, int target) {
 int[] sumnums = nums.clone();
 //int[2] sol = {0,1};
 int[] sol = new int[]{0,1};

 sol = new int[2];
 int j=sumnums.length; // Every recursion j will be initialized as sumnums.length instead of J-1
 int t=target;

 for(int i=0;i&lt;sumnums.length;i++){
 // if ((sumnums[i]+sumnums[j])==t){ 
 // Remember that Java arrays start at index 0, so this value(sumnums[j]) is not exist in the array.
 if ((sumnums[i]+sumnums[j-1])==t){
 sol[0]=i;
 sol[1]=j;
 return sol;
 }
 }
 j=j-1;
 twoSum(sumnums,t);
 return sol;

 }
</code></pre>

blocks|key|4109547|text|我使用更简单的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4109548|类解决方案{+public+int[]+twoSum(int[]数字，整型目标){|4109549|+++++int[]+indices=+new+int[2];
++++int+len=nums.length;
++++for(int+i=0;+i<len;+i%2B%2B){

++++++++++++for(int+j=i%2B1;+j<len;j%2B%2B){+++++++++

++++++++++++++++if((nums[i]%2Bnums[j])==target){
++++++++++++++++++++indices[0]+=+i;
++++++++++++++++++++indices[1]+=+j;
++++++++++++++++++++return+indices;

++++++++++++++++}
++++++++++++}


++++}
+++++return+null;
}|code-block|syntax|javascript|4109550|}|4109551|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I use the simpler approch

class Solution {
 public int[] twoSum(int[] nums, int target) {

<pre><code> int[] indices= new int[2];
 int len=nums.length;
 for(int i=0; i&lt;len; i++){

 for(int j=i+1; j&lt;len;j++){ 

 if((nums[i]+nums[j])==target){
 indices[0] = i;
 indices[1] = j;
 return indices;

 }
 }


 }
 return null;
}
</code></pre>

}

blocks|key|4109283|text|为什么不使用HashMap呢？下面是它的工作原理。我们将遍历数组并检查映射中是否存在target+-+nums[i]。如果不是，我们将Key,+Value分别存储为数字和它的索引，即K+=+nums[i],+V+=+i|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4109284|下面是它的工作原理:考虑一下nums+=+[2,+7,+11,+15],+target+=+9|4109285|我们将开始迭代数组|4109286|首先是2。这里我们将检查(9-2)，即7不存在于hashmap中，因此我们将存储2作为key，并将其索引0作为其值|4109287|这里我们将检查映射中存在的(9-+7+)即2，因此我们将返回索引2和索引7，即返回0，1|4109288|public+int[]+twoSum(int[]+nums,+int+target)+{

++++++++HashMap<Integer,Integer>+map+=+new+HashMap<>();
+++++++++for(int+i+=+0;+i+<+nums.length;+i%2B%2B)+
+++++++++{
+++++++++++++if(map.containsKey(target+-+nums[i]))+
+++++++++++++{
+++++++++++++++++return+new+int[]+{map.get(target-nums[i]),i};

+++++++++++++}
+++++++++++++else
+++++++++++++++++map.put(nums[i],i);
+++++++++}
+++++++++return+new+int[]+{-1,-1};
++++}|code-block|syntax|javascript|4109289|entityMap^0|16|G|1V|A|2J|I|0|E|X|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|14|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|15|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|16|8|@]|D|@]|E|$]]|$1|N|3|O|5|P|7|17|8|@]|D|@]|E|$Q|R]]|$1|S|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|T|$]]

Why not use a HashMap? Here is how it will work. We will iterate through the array and check if the <code>target - nums[i]</code> exists in the map. If not, we will store <code>Key, Value</code> as the number and its index respectively i.e <code>K = nums[i], V = i</code>

Here is how it will work: 
Consider <code>nums = [2, 7, 11, 15], target = 9</code>

We will start iterating the array

First comes 2. Here we will check ( 9 - 2 ) i.e 7 does not exist in the hashmap, so we will store 2 as key and its index 0 as its value

Then, comes 7. Here we will check ( 9 - 7 ) i.e 2 which exists in the map and so we will return the index of 2 and index of 7 i.e returning [ 0, 1 ]

<pre><code>public int[] twoSum(int[] nums, int target) {

 HashMap&lt;Integer,Integer&gt; map = new HashMap&lt;&gt;();
 for(int i = 0; i &lt; nums.length; i++) 
 {
 if(map.containsKey(target - nums[i])) 
 {
 return new int[] {map.get(target-nums[i]),i};

 }
 else
 map.put(nums[i],i);
 }
 return new int[] {-1,-1};
 }
</code></pre>

I am attempting a Java mock interview on LeetCode. I have the following problem:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

<pre><code>Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
</code></pre>

I was attempting to implement a recursive solution. However, I am receiving errors upon trying to run my code.

<pre><code>class Solution {
 public int[] twoSum(int[] nums, int target) {
 int[] sumnums = nums.clone();
 //int[2] sol = {0,1};
 int[] sol = new int[]{0,1};

 sol = new int[2];
 int j=sumnums.length;
 int t=target;

 for(int i=0;i&lt;sumnums.length;i++){
 if ((sumnums[i]+sumnums[j])==t){
 sol[0]=i;
 sol[1]=j;
 //return sol;
 }
 }
 j=j-1;
 twoSum(sumnums,t);
 return sol;

 }
}
</code></pre>

Error(s): 

<pre><code>Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 4
 at Solution.twoSum(Solution.java:12)
 at __DriverSolution__.__helper__(__Driver__.java:8)
 at __Driver__.main(__Driver__.java:54)
</code></pre>

It appears to me the error may have to do with the following line of code:

<pre><code>if ((sumnums[i]+sumnums[j])==t){
</code></pre>

Therefore, I am wondering if this is a syntax related error. I am attempting to check to see if two numbers add up to a different number.

Since this is a naive attempt at a recursive solution, I am happy to take any other criticism. But I am mostly concerned with getting my attempt at this problem to work and run with all testcases.

Thanks.

Two Sum LeetCode Java Questions

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正尝试在LeetCode上进行一次Java模拟面试。我有以下问题：给定一个整数数组，返回这两个数字的索引，使它们相加为一个特定的目标。您可以假设每个输入都只有一个解决方案，并且不能使用同一元素两次。Example:Given nums = [2, 7, 11, 15], target = 9,Because nums...

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