给两个排序好的整数数组，数组的长度是相同的，找到这两个数组的上中位数，也就是如果数组是偶数的话，输出前一个中位数。

时间复杂度O（logN），空间复杂度O（1）

public class UpMedian{        public static int getUpMedian(int[] arr1,int[] arr2) {        if(arr1==null || arr1.length<=0 || arr2==null || arr2.length<=0) {            System.out.println("Array is valid");            return -1;        }                if(arr1.length!=arr2.length) {            System.out.println("Array length is valid");            return -1;        }                int start1 = 0;        int end1 = arr1.length-1;        int middle1 = 0;        int start2 = 0;        int end2 = arr2.length-1;        int middle2 =0;        //用来区分数组长度为奇偶数        int offset = 0;                while(start1 < end1) {            middle1 = (start1+end1)>>1;            middle2 = (start2+end2)>>1;            offset = ((end1-start1+1)&1)^1;                        if(arr1[middle1] > arr2[middle2]) {                end1=middle1;                start2=middle2+offset;            } else if(arr1[middle1] < arr2[middle2]) {                start1 = middle1 +offset;                end2 = middle2;            } else {                return arr1[middle1];            }        }                return Math.min(arr1[start1], arr2[start2]);    }    public static void main(String[] args) {        int[] a1 = {1,2,5,7};        int[] a2 = {2,3,8,10};        System.out.println(getUpMedian(a1, a2));    }}

这种题重在分析各种情况，

基本原理就是二分查找，但是你要确定你查找最合适的子数组，这样才能达到O（logN）的时间复杂度