二分查詢變形問題：

（1）查詢第一個值等於給定值的元素

（2）查詢最後一個值等於給定值的元素

（3）查詢第一個大於等於給定值的元素

（4）查詢最後一個小於等於給定值的元素

//（1）查詢第一個值等於給定值的元素
public int bsearch1(int[] a, int n, int value){
    int low = 0;
    int high = n-1;
    while (low<=high){
        int mid = low + ((high - low) >> 1);
        if (a[mid] > value){
            high = mid - 1;
        }else if (a[mid]<value){
            low = mid + 1
        }else{
            if ((mid == 0) || (a[mid-1] != value)) return mid;
            else high = mid - 1;
        }
    }
    return -1;
}

//（2）查詢最後一個值等於給定值的元素
public int bsearch2(int[] a, int n, int value){
    int low = 0;
    int high = n-1;
    while (low<=high){
        int mid = low + ((high - low) >> 1);
        if (a[mid] > value){
            high = mid - 1;
        }else if (a[mid]<value){
            low = mid + 1
        }else{
            if ((mid == n-1) || (a[mid+1] != value)) return mid;
            else low = mid + 1;
        }
    }
    return -1;
}

//（3）查詢第一個大於等於給定值的元素
public int bsearch3(int[] a, int n, int value){
    int low = 0;
    int high = n-1;
    while (low<=high){
        int mid = low + ((high - low) >> 1);
        if (a[mid] >= value){
            if ((mid==0 || a[mid-1] < value)) return mid;
            else high = mid - 1;
        }else{
            low = mid + 1
        }
    }
    return -1;
}

//（4）查詢最後一個小於等於給定值的元素
public int bsearch4(int[] a, int n, int value){
    int low = 0;
    int high = n-1;
    while (low<=high){
        int mid = low + ((high - low) >> 1);
        if (a[mid] > value){
            high = mid - 1; 
        }else {
            if ((mid==n-1 || a[mid+1] > value)) return mid;
            else low = mid + 1;
        }
    }
    return -1;
}