从重复数组中查找丢失元素的 Javascript 程序

原文:https://www . geesforgeks . org/JavaScript-program-for-find-lost-element-from-a-replicated-array/

假设两个数组除了一个元素之外都是重复的,也就是说其中一个数组的一个元素丢失了,我们需要找到那个丢失的元素。 例:

Input:  arr1[] = {1, 4, 5, 7, 9}
        arr2[] = {4, 5, 7, 9}
Output: 1
1 is missing from second array.

Input: arr1[] = {2, 3, 4, 5}
       arr2[] = {2, 3, 4, 5, 6}
Output: 6
6 is missing from first array.

一个简单的解决方案是迭代数组,逐元素检查,当发现不匹配的元素时标记缺失的元素,但是这个解决方案需要数组的线性时间过大。 另一个高效解决方案是基于 一个二分搜索法的方法。算法步骤如下:

  1. 在更大的阵列中启动二分搜索法,获得中间 as (lo + hi) / 2
  2. 如果两个数组的值相同,那么缺少的元素必须在右边,因此将 lo 设置为 mid
  3. 否则将 hi 设置为 mid,因为如果 mid 元素不相等,丢失的元素必须在更大数组的左边。
  4. 一种特殊情况是单独处理的,对于单元素和零元素数组,单元素本身将是缺失的元素。 如果第一个元素本身不相等,那么该元素将是缺失的元素。/li >

以下是上述步骤的实施

java 描述语言

<script>
    // Javascript program to find missing element from
    // same arrays (except one missing element)

    /* Function to find missing element based
    on binary search approach. arr1[] is of
    larger size and N is size of it.arr1[] and
    arr2[] are assumed to be in same order. */
    function findMissingUtil(arr1, arr2, N)
    {

        // special case, for only element
        // which is missing in second array
        if (N == 1)
            return arr1[0];

        // special case, for first
        // element missing
        if (arr1[0] != arr2[0])
            return arr1[0];

        // Initialize current corner points
        let lo = 0, hi = N - 1;

        // loop until lo < hi
        while (lo < hi) {
            let mid = parseInt((lo + hi) / 2, 10);

            // If element at mid indices are
            // equal then go to right subarray
            if (arr1[mid] == arr2[mid])
                lo = mid;
            else
                hi = mid;

            // if lo, hi becomes
            // contiguous, break
            if (lo == hi - 1)
                break;
        }

        // missing element will be at hi
        // index of bigger array
        return arr1[hi];
    }

    // This function mainly does basic error
    // checking and calls findMissingUtil
    function findMissing(arr1, arr2, M, N)
    {
        if (N == M - 1)
            document.write("Missing Element is "
            + findMissingUtil(arr1, arr2, M) + "</br>");
        else if (M == N - 1)
            document.write("Missing Element is "
            + findMissingUtil(arr2, arr1, N) + "</br>");
        else
            document.write("Invalid Input" + "</br>");
    }

    let arr1 = [ 1, 4, 5, 7, 9 ];
    let arr2 = [ 4, 5, 7, 9 ];
    let M = arr1.length;
    let N = arr2.length;
    findMissing(arr1, arr2, M, N);

</script>

输出:

Missing Element is 1

如果输入数组不在 中,该怎么办? 在这种情况下,缺失的元素只是两个数组所有元素的异或。感谢 Yolo Song 的建议。

java 描述语言

<script>

    // Javascript program to find 
    // missing element from one array
    // such that it has all elements 
    // of other array except one.
    // Elements in two arrays
    // can be in any order.

    // This function mainly does XOR 
    // of all elements
    // of arr1[] and arr2[]
    function findMissing(arr1, arr2, M, N)
    {
        if (M != N-1 && N != M-1)
        {
            document.write("Invalid Input");
            return;
        }

        // Do XOR of all element
        let res = 0;
        for (let i=0; i<M; i++)
           res = res^arr1[i];
        for (let i=0; i<N; i++)
           res = res^arr2[i];

        document.write("Missing element is " + res);
    }

    let arr1 = [4, 1, 5, 9, 7];
    let arr2 = [7, 5, 9, 4];

    let M = arr1.length;
    let N = arr2.length;

    findMissing(arr1, arr2, M, N);

</script>

输出:

Missing Element is 1

更多详细信息,请参考完整文章从重复数组中查找丢失元素!