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) ;

        }

    }

巴特西

Arrays.binarySearch采坑记录及用法

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