Median of Two Sorted Arrays

最新推荐文章于 2023-10-07 04:23:19 发布

yongdong_wang

最新推荐文章于 2023-10-07 04:23:19 发布

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分类专栏： Leetcode 文章标签： Leetcode Java Algorithm

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本文介绍了一种算法来查找两个已排序数组的中位数，整体运行复杂度为O(log(m+n))。该算法通过合并数组并找到中间元素来实现目标。

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There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

//下面这个算法只做到了O(m+n),虽然还是被oj接受了，但还是要继续改进

int[]   merge(int[] arrToBeMergedA, int[] arrToBeMergedB){
    if(arrToBeMergedA == null || arrToBeMergedB == null){

        return null;

    }else if(arrToBeMergedA == null){

        return arrToBeMergedB;

    }else if(arrToBeMergedB == null){

        return arrToBeMergedA;  

    }else {
        int[]   mergedArr = new int[arrToBeMergedA.length + arrToBeMergedB.length];
        int pointA = 0;
        int pointB = 0;
        for(int i  = 0; i < mergedArr.length; ){
            if(pointA < arrToBeMergedA.length && pointB < arrToBeMergedB.length){
                mergedArr[i++] = arrToBeMergedA[pointA]<=arrToBeMergedB[pointB]?arrToBeMergedA[pointA++]:arrToBeMergedB[pointB++];
            }else if(pointA >= arrToBeMergedA.length){
                mergedArr[i++] = arrToBeMergedB[pointB++];
            }else{
                mergedArr[i++] = arrToBeMergedA[pointA++];
            }
        }
        return mergedArr;
    }
}

    public double findMedianSortedArrays(int A[], int B[]) {
     int[] myArr = merge(A,B);
     int middle = myArr.length / 2;
     double median = 0;
     if(myArr.length % 2 == 0){
        median = (myArr[middle] + myArr[middle-1]) / 2.0;
     }else{
        median = myArr[middle];
     }
     return median;
    }