查找中间数

package com.viking.divide;/** *  * @author viking *  *         查找中间数 有两个长度相等，按升序排列的数组，现要查找中间数 因为有两个中间数，返回偏小的那一个 *  *         中间数是数组中大小处于中间的那个数 *  *         基本思路，用而分查找的方法查找 *  *  *  */public class Middle {public static void main(String[] args) {int[] a = {15,  16, 23,27 };int[] b = {7, 10, 12,13 };int mid = findMiddle(a, b);System.out.println(mid);}public static int findMiddle(int[] a, int[] b) {int n = a.length;int amid = 0, bmid = 0;int alow = 0, blow = 0;int ahight = n - 1, bhight = n - 1;while (alow < ahight && blow < bhight) {amid = (alow + ahight) / 2;bmid = (blow + bhight) / 2;if (a[amid] > b[bmid]) {// 中间数会在 a的中 alow-amid半区和b的blow-bhight半区中ahight = amid;blow = bmid;} else {// 中间数会在 b的中 blow-bmid半区和a的alow-ahight半区中alow = amid;bhight = bmid;}}// 由于取整问题，下标总是指向偏小的数// a[amid+1]或者b[bmid+1]有可能是中间数// middle 肯定在a[amid] a[amid+1] b[bmid] b[bmid+1]这四个数中// 对amid bmid 进行修正if (amid + bmid + 1 == n - 2) {if (a[amid] > b[bmid]) {amid++;} else {bmid++;}} else if (amid + bmid + 2 == n - 2) {amid++;bmid++;}if (a[amid] > b[bmid]) {if (bmid < n && a[amid] > b[bmid + 1]) {// bmid+1才是偏小的那个中间数return b[bmid + 1];}return a[amid];} else {if (amid < n && b[bmid] > a[amid + 1]) {// amid+1才是偏小的那个中间数return a[amid + 1];}return b[bmid];}}}