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
 * 给定两个大小分别为 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 。
 *
 * 算法的时间复杂度应该为 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;
        //临时存储K
        int kTemp = k;
        //记录遍历次数
        int count = 0;
        while (true) {
            //特殊情况，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];
            }
            //已经遍历了 k-1次 等同于 已经找到了目标元素
            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，谁移动指针
            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;

            }
        }
    }