再探二分查找

public class BiSearchChangeMode1{
    public static void main(String args[]){
        double[] a = {1,2.5,2.5,2.5,2.5,2.5,3,4,4,4,5};
        int len = a.length;
        double key = 25; //四组测试数据(-2,2,2.5,25)
        
        int resIndex = getTheMinThanKey(a,len,key,0,len-1);
        System.out.println("+++++++++++++++++++++++++++++++++++");
        System.out.println("1.在一个有序的数组中,找到一个最小的比key大的数(即,第一个比key大的数)");
        if(resIndex != -1){
            System.out.println("\t数组的下标为:" + resIndex);
            System.out.println("\t数组的值为:" + a[resIndex]);
        }else{
            System.out.println("\t数组中没有满足条件的值");
        }
                
        System.out.println("+++++++++++++++++++++++++++++++++++");
        System.out.println("2.在一个有序的数组中,找到和key相等的最小下标的数(即,第一个和key相等的数)");
        resIndex = getTheFirstEqualsNum(a,len,key,0,len-1);
        if(resIndex != -1){
            System.out.println("\t数组的下标为:" + resIndex);
            System.out.println("\t数组的值为:" + a[resIndex]);
        }else{
            System.out.println("\t数组中没有满足条件的值");
        }
        
        System.out.println("+++++++++++++++++++++++++++++++++++");
        System.out.println("3.在一个有序的数组中,找到和key相等的最大下标的数(即,最后一个和key相等的数)");
        resIndex = getTheLastEqualsNum(a,len,key,0,len-1);
        if(resIndex != -1){
            System.out.println("\t数组的下标为:" + resIndex);
            System.out.println("\t数组的值为:" + a[resIndex]);
        }else{
            System.out.println("\t数组中没有满足条件的值");
        }
        
        System.out.println("+++++++++++++++++++++++++++++++++++");
        System.out.println("4.在一个有序的数组中,找到和key相等的数(任意一个),不存在则返回-1");
        resIndex = getAnyEqualsNum(a,key,0,len-1);
        if(resIndex != -1){
            System.out.println("\t数组的下标为:" + resIndex);
            System.out.println("\t数组的值为:" + a[resIndex]);
        }else{
            System.out.println("\t数组中没有满足条件的值");
        }

        System.out.println("+++++++++++++++++++++++++++++++++++");
        System.out.println("5.在一个有序的数组中,找到一个最大的比key小的数(即,最后一个比key小的数)");
        resIndex = getTheMaxLessKey(a,len,key,0,len-1);
        if(resIndex != -1){
            System.out.println("\t数组的下标为:" + resIndex);
            System.out.println("\t数组的值为:" + a[resIndex]);
        }else{
            System.out.println("\t数组中没有满足条件的值");
        }

    }

    /**
    * 1 在一个有序的数组中,找到一个最小的比key大的数(即,第一个比key大的数)
    */
    public static int getTheMinThanKey(double[] a,int len,double key,int low,int high){
        //1:如果第一个数就比key大
        if(a[0]>key){
            return 0;
        }
        //2:如果最后一个数不大于key
        if(a[len-1]<=key){
            return -1;
        }
        //3 :其他情况
        if(high>=1 && a[high]>key && a[high-1]<=key){
            return high;
        }
        while(low < high-1){
            int mid = low + (high-low)/2;
            if(a[mid]>key)
                high = mid;
            else 
                low = mid;
        }
        return getTheMinThanKey(a,len,key,low,high);
    }


    /**
    * 2 在一个有序的数组中,找到和key相等的最小下标的数(即,第一个和key相等的数)
    */
    public static int getTheFirstEqualsNum(double[] a,int len,double key,int low,int high){
        //1:没有满足条件的
        if(low>high){
            return -1;
        }
        //2:第一个数即满足条件时
        if(a[0] == key){
            return 0;
        }
        //3:其他情况
        if(high>=1 && a[high]==key && a[high-1] != key){
            return high;
        }
        while(low<=high){
            int mid = low + (high-low)/2;
            if(a[mid]>key)
                high = mid-1;
            else if(a[mid]<key)
                low = mid+1;
            else{
                high = mid;
                break;
            }
        }
        return getTheFirstEqualsNum(a,len,key,low,high);
    }

    /**
    * 3 在一个有序的数组中,找到和key相等的最大下标的数(即,最后一个和key相等的数)
    */
    public static int getTheLastEqualsNum(double[] a,int len,double key,int low,int high){
        //1:没有满足条件的
        if(low>high){
            return -1;
        }
        //2:最后一个数即满足条件时
        if(a[len-1] == key){
            return len-1;
        }
        //3:其他情况
        if(low<len-1 && a[low]==key && a[low+1] != key){
            return low;
        }
        while(low<=high){
            int mid = low + (high-low)/2;
            if(a[mid]>key)
                high = mid-1;
            else if(a[mid]<key)
                low = mid+1;
            else{
                low = mid;
                break;
            }
        }
        return getTheLastEqualsNum(a,len,key,low,high);
    }

    /**
    * 4 在一个有序的数组中,找到和key相等的数(任意一个),不存在则返回-1
    */
    public static int getAnyEqualsNum(double[] a,double key,int low,int high){
        while(low<=high){
            int mid = low + (high-low)/2;
            if(a[mid]>key){
                high = mid - 1;
            }else if(a[mid]<key){
                low = mid + 1;
            }else{
                return mid;
            }
        }
        return -1;
    }

    /**
    * 5 在一个有序的数组中,找到一个最大的比key小的数(即,最后一个比key小的数)
    */
    public static int getTheMaxLessKey(double[] a,int len,double key,int low,int high){
        //1:如果最后一个数比key小
        if(a[len-1]<key){
            return len-1;
        }
        //2:如果第一个比key大
        if(a[0]>key){
            return -1;
        }
        if(low<len-1 && a[low]<key && a[high]>=key){
            return low;
        }
        while(low<high-1){
            int mid = low + (high-low)/2;
            if(a[mid]>key){
                high = mid;
            }else{
                low = mid;
            }
        }
        return getTheMaxLessKey(a,len,key,low,high);
    }
}

 

posted @ 2014-10-30 09:03  NW_KNIFE  阅读(232)  评论(0编辑  收藏  举报