核心思路：游标记录已经遍历到当前位置

时间复杂度：m+n(两个数组的长度)

public class MergeDemo {    /**     * @param args     */    public static void main(String[] args) {        int[] a = {1,4,6,8};        int[] b = {3,5,7,10,20};        merger(a,b);                int[] c = {1,2,3,4,5};        int[] d = {11,12,13,14,15};        merger(d,c);                int[] e = {1,12,13,14,15};        int[] f = {11,12,13,14,15};        merger(e,f);    }        public static void merger(int[] a,int[] b){        int[] c = new int[a.length+b.length];        int indexA = 0;        int indexB = 0;        for(int i=0;i
     
      =a.length&&indexB
      
       =b.length&&indexA