二分查找，数组去重Java

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/** * 二分查找 */package algorithms;/** * @author  * * @time 2017年5月9日 下午3:48:29 */public class BinarySearch {    public static int rank(int key, int[] a){        //数组必须是有序的        int head = 0;        int tail = a.length - 1;        while(head <= tail){            //被查找的键要么不存在，要么必然存在于a[head..tail]之中            int mid = head + (tail - head)/2;            if(key < a[mid]){                tail = mid - 1;            }else if(key > a[mid]){                head = mid + 1;            }else{                return mid;            }        }        return -1;    }    //递归法    public static int binarySearch(int[] a,int head,int tail,int key){        //数组必须是有序的        //        int head = 0;        //        int tail = a.length - 1;        if(head <= tail){            //被查找的键要么不存在，要么必然存在于a[head..tail]之中            int mid = head + (tail - head)/2;            if(key < a[mid]){                tail = mid - 1;            }else if(key > a[mid]){                head = mid + 1;            }else{//key == a[mid]                return mid;            }            return binarySearch(a,head,tail,key);        }        return -1;    }    public static void main(String[] args) {//        int[] a = {10,20,30,40,50,60};//        int i = rank(40,a);//        System.out.println(i);//        if(i>-1){//            System.out.println(a[i]);//        }//        System.out.println(binarySearch(a,0,a.length-1,40));        int[] a = {10,20,30,30,30,40,40,40,40,50,60};        int[] b = duplicateRemoval(a);//去重后的数组    }  /**     * 数组去重     */    private static int[] duplicateRemoval(int[] a) {        //        int num = count(a);        int[] b = new int[a.length-num];        int count = 0;        b[0] = a[0];        for(int i = 0;i < a.length-1;i++){            if(a[i]==a[i+1]){                count ++;            }else{                b[i+1-count] = a[i+1];            }        }        return b;    }    /**     * 计算重复数量      */    private static int count(int[] a) {        int count = 0;        for(int i = 0;i < a.length-1;i++){            if(a[i]==a[i+1]){                count ++;            }        }        return count;    }}
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