关于算法:4寻找两个正序数组的中位数

/** * 4.寻找两个正序数组的中位数 * 给定两个大小别离为 m 和 n 的正序（从小到大）数组nums1 和nums2。请你找出并返回这两个正序数组的 中位数 。 * * 算法的工夫复杂度应该为 O(log (m+n)) * */public class Median {    public static double findMedianElement(int[] numArray1, int[] numArray2) {        int length1 = numArray1.length, length2 = numArray2.length;        int totalLength = length1 + length2;        //奇数        if (totalLength % 2 == 1) {            int midIndex = totalLength / 2;            double median = getElement(numArray1, numArray2, midIndex + 1);            return median;        } else {            //偶数            int midIndex1 = totalLength / 2 - 1, midIndex2 = totalLength / 2;            int k1 = getElement(numArray1, numArray2, midIndex1 + 1);            int k2 = getElement(numArray1, numArray2, midIndex2 + 1);            double median = ( k1 + k2 ) / 2.0;            return median;        }    }    public static int getElement(int[] numArray1, int[] numArray2, int k) {        int length1 = numArray1.length, length2 = numArray2.length;        int index1 = 0, index2 = 0;        //长期存储K        int kTemp = k;        //记录遍历次数        int count = 0;        while (true) {            //非凡状况，取第一个元素            if (k == 1) {                if (length1 == 0){                    return numArray2[index2];                }                if (length2 == 0){                    return numArray1[index1];                }                return Math.min(numArray1[index1], numArray2[index2]);            }            //元素初始值            int pivot1 = -99999, pivot2 = -99999;            //数组1，下标没有越界，且数组个数大于0            if (index1 <= length1 -1 && length1 > 0){                pivot1 = numArray1[index1];            }            //数组2，下标没有越界，且数组个数大于0            if (index2 <= length2 -1 && length2 > 0){                pivot2 = numArray2[index2];            }            //曾经遍历了 k-1次 等同于 曾经找到了指标元素            if(count == (kTemp-1)){                if (index1 > length1 - 1) {                    return pivot2;                }                if (index2 > length2 - 1) {                    return pivot1;                }                return Math.min(pivot1,pivot2);            }            //谁的元素小，谁挪动指针            if ((pivot1 != -99999 && pivot2 != -99999 && pivot1 <= pivot2) ||                    pivot2 == -99999            ) {                //右移 nums1 指针                index1 += 1;                //记录遍历次数                count += 1;            } else {                //右移 nums2 指针                index2 += 1;                //记录遍历次数                count += 1;            }        }    }