Arrays.binarySearch采坑记录及用法

今天在生产环境联调的时候,发现一个很奇怪的问题,明明测试数据正确,结果却是结果不通过,经过debug查询到原来是Arrays.binarySearch用法错误,记录一下,避免后续再次犯错

具体测试如下:

想通过判断J是否存在数组中,结果发现出现如下错误

 public static void main(String[] args) {
        String[] test ={"X","J","7","5","4","11","W8","W7"};
        System.out.println("没排序结果 = [" + Arrays.binarySearch(test,"J") + "]");
    }

运行结果:

没有排序 = [-7]

进行排序后判断:

 public static void main(String[] args) {
        String[] test ={"X","J","7","5","4","11","W8","W7"};
        System.out.println("没排序结果 = [" + Arrays.binarySearch(test,"J") + "]");
        Arrays.sort(test);
        System.out.println("排序后结果 = [" + Arrays.binarySearch(test,"J") + "]");
    }

运行结果:

没排序结果 = [-7]
排序后结果 = [4]

在网上查询下具体原因及查询官方解释如下:

/**
     * Searches the specified array for the specified object using the binary
     * search algorithm. The array must be sorted into ascending order
     * according to the
     * {@linkplain Comparable natural ordering}
     * of its elements (as by the
     * {@link #sort(Object[])} method) prior to making this call.
     * If it is not sorted, the results are undefined.
     * (If the array contains elements that are not mutually comparable (for
     * example, strings and integers), it <i>cannot</i> be sorted according
     * to the natural ordering of its elements, hence results are undefined.)
     * If the array contains multiple
     * elements equal to the specified object, there is no guarantee which
     * one will be found.
     *
     * @param a the array to be searched
     * @param key the value to be searched for
     * @return index of the search key, if it is contained in the array;
     *         otherwise, <tt>(-(<i>insertion point</i>) - 1)</tt>.  The
     *         <i>insertion point</i> is defined as the point at which the
     *         key would be inserted into the array: the index of the first
     *         element greater than the key, or <tt>a.length</tt> if all
     *         elements in the array are less than the specified key.  Note
     *         that this guarantees that the return value will be >= 0 if
     *         and only if the key is found.
     * @throws ClassCastException if the search key is not comparable to the
     *         elements of the array.
     */
    public static int binarySearch(Object[] a, Object key) {
        return binarySearch0(a, 0, a.length, key);
    }

使用二分搜索法来搜索指定数组，以获得指定对象。在进行此调用之前，必须根据元素的自然顺序对数组进行升序排序（通过 sort(Object[]) 方法）。如果没有对数组进行排序，则结果是不确定的。（如果数组包含不可相互比较的元素（例如，字符串和整数），则无法 根据其元素的自然顺序对数组进行排序，因此结果是不确定的。）如果数组包含多个等于指定对象的元素，则无法保证找到的是哪一个,故所以会出现此问题

如果要判断数组中是否存在,可以使用 ArrayUtils.contains这个方法来判断是否存在,

 

/**
     * 校验服务类型是否符合枚举值
     * @param possibility 可能性
     * @param field 字段
     * @param info 校验不通过的错误消息
     *
     */
    public void validatePossibility(String[] possibility,String field,ErrorInfo info){
        boolean flag = ArrayUtils.contains(possibility,field);
        if (!flag) {
            throw getException(info) ;
        }
    }