如果数组中有任意两个元素(I +j)等于给定的键，该如何求?

Let's say I am given an array

假设我有一个数组

int[] arr = {1, 2, 3, 4, 5, 6}

and I have to input a "key" value which is based on user input.

我必须输入一个基于用户输入的“键”值。

If the array have any two elements that sums up to my key value it returns a true value or otherwise a false value. Why do I still have issue with my algorithm? Is there some misconception that I am putting in the nested loop? Any help with this please?

如果数组中有任何两个元素总和为我的键值，它将返回一个true值或一个false值。为什么我的算法仍然有问题?是否存在将我放入嵌套循环的误解?有什么帮助吗?

public static boolean checkPair(int[] arr, int key) {
    for (int i = 0; i < arr.length; i++) {
      for (int j = i + 1; j < arr.length; j++) {
        if (arr[i] + arr[j] == key) {
          return true;
        } else {
          return false;
        }
      }
    }
    return false;  
}

3 个解决方案

#1

Your else-clause is wrong. It always returns false if the first pair hasn't the same value.

你的else子句是错误的。如果第一对的值不相同，则返回false。

Change your code like this:

像这样更改代码:

public static boolean checkPair(int[] arr, int key) {
    for (int i = 0; i < arr.length; i++) {
       for (int j = i + 1; j < arr.length; j++) {
         if (arr[i] + arr[j] == key) {
           return true;
         }
        //else {      
        //  return false;    <- this returns always false if the 
        //                      first pair isnt a i,j-match
        //}
      } 
    }
    return false;  // stub
}

Explanation

解释

In your example int[] arr = {1, 2, 3, 4, 5, 6}

在您的示例int[] arr = {1,2,3,4,5,6}

arr[i] = 1 and a[j] = 2 If you set key to any int except 3, the method will return false and the other numbers will not be considered

arr[i] = 1, a[j] = 2如果你将键设置为除3之外的任意整数，该方法将返回false，其他数字将不被考虑

#2

I think, this task can be solved faster than your algorithm. Are your array sorted?? If yes, you can do that:

我认为，这个任务可以比你的算法更快地解决。你的数组排序?如果是，你可以这样做:

public static boolean checkPair(int[] arr, int key) {
  int i = 0; //first index
  int j = arr.length - 1; //last index
  while (i < j) {
    if (arr[i] + arr[j] > key){
      j--;
      continue;
    }
    if (arr[i] + arr[j] < key){
      i++;
      continue;
    }
    else {
      return true;
    }
  }
  return false;
}

If your array are not sorted, you can sort it with java standart sort and use my algorithm. My algorithm faster more than 2 times.

如果数组没有排序，可以使用java standart排序并使用我的算法。我的算法快了2倍多。

#3

-2

you just break loop like this:

你只需要像这样打破循环:

public static boolean checkPair(int[] arr, int key) {

    boolean isMacthed = false;
    for (int i=0; i<arr.length; i++) {
      for (int j=i+1; j<arr.length; j++) {
        if (arr[i] + arr[j] == key) {
          isMacthed=true;
          break;
        }
      }
      if(isMacthed) break;
    }
    return isMacthed ; 
    }

#1