JAVA二分查找

请高手写个 JAVA二分查找 ~请指教~

public class test {
static int bsearch( int[] a, int v ) {
int l, r;
l = 0; r = a.length-1;
while ( l <= r ) {
int m = (l+r)/2;
if ( a[m] == v ) return m; else
if ( a[m] > v ) r = m-1; else
if ( a[m] < v ) l = m+1;
}
return -1;
}

public static void main( String[] args ) {
int[] a = { 1,3,5,7,9 };
for ( int i = 0; i < 10; ++i )
System.out.println( "find " + i + " at pos: " + bsearch( a, i ) );
}
}

以下代码是关于对象的 二分查找 的例子，已经测试通过，执行即可。 student 是基本比较对象类 dichotomy 是二分法执行类 test 是测试类 package com.dichotomy; public class student implements comparable { private int id; private string name; private string idcard; private string sex; private string mobile; public int getid() { return id; } public void setid(int id) { this.id = id; } public string getname() { return name; } public void setname(string name) { this.name = name; } public string getidcard() { return idcard; } public void setidcard(string idcard) { this.idcard = idcard; } public string getsex() { return sex; } public void setsex(string sex) { this.sex = sex; } public string getmobile() { return mobile; } public void setmobile(string mobile) { this.mobile = mobile; } public int compareto(student o) { //不同的执行次序决定排序和查找次序不同,可以同下面的调换一下 if(this.getid() < o.getid()){ return -1; } else if(this.getid() == o.getid()){ ; } else { return 1; } //不同的执行次序决定排序和查找次序不同 int c = this.getidcard().compareto(o.getidcard()); if(c != 0){ return c; } //不同的执行次序决定排序和查找次序不同 int n = this.getname().compareto(o.getname()); if(n != 0){ return n; } return 0; } public string tostring(){ stringbuffer sb = new stringbuffer(); sb.append(this.getid()).append("\t"); sb.append(this.getname()).append("\t"); sb.append(this.getidcard()).append("\t"); sb.append(this.getmobile()).append("\t"); sb.append(this.getsex()); return sb.tostring(); } }