二分搜索错误情况的分析

又到了一年一季的招聘季，二分搜索是一个烂大街的题目，相信会写的人不少，当你能真正知道错误写法到底错在哪里吗？

错误一，因结束条件判断错误，数组中明明存在key，但程序却却找不到

public static int binary_search(int [] array, int key, int begin, int end) {
        int mid;
        while(begin<end) {
            mid = begin + (end - begin)/2;
            System.out.println("begin = " + begin + " mid = " + mid + " end = " + end);
            if (array[mid]==key) {
                return mid;
            }
        
            if(array[mid]>key) {
                end = mid -1;
            }
            else {
                begin = mid + 1;
            }
        }
        return 0;
}

测试用的代码：

         int array[] = {0, 1, 2, 3, 4, 5, 6, 7, 13, 19};
         int index = BinarySearch.search(array, 4, 0, 9);
        System.out.println("index = " + index);

执行的结果：

        begin = 0 mid = 5 end = 10
        begin = 0 mid = 2 end = 4
        begin = 3 mid = 3 end = 4
        index  = 0

你会发现明明数组中存在4，但却没有找到。原因就是判断程序结束时使用了：左闭右开的区间:[0, n)。 因此，当这个条件满足时，end应该为mid （已判断要搜索的key不是mid）， 而在这里right赋值为middle - 1了, 判断区间为：[0, middle-1)，那么就会遗漏array[middle - 1] = v的情况。

错误二，无限循环

    public static int binary_search (int [] array, int key, int begin, int end) {
        int mid;
        while(begin<=end) {
            mid = begin + (end - begin)/2;
            System.out.println("begin = " + begin + " mid = " + mid + " end = " + end);
            if (array[mid]==key) {
                return mid;
            }
        
            if(array[mid]>key) {
                end = mid;
            }
            else {
                begin = mid;
            }
        }
        return 0;
    }

 int array[] = {1, 2, 3, 3, 5, 7, 8, 13, 15,19};
 int index = BinarySearch.search_1(array, 4, 0, 9);

 执行的结果：

    begin = 0 mid = 4 end = 9
    begin = 0 mid = 2 end = 4
    begin = 2 mid = 3 end = 4 
    begin = 3 mid = 3 end = 4
    begin = 3 mid = 3 end = 4
    begin = 3 mid = 3 end = 4
    begin = 3 mid = 3 end = 4 

无限循环

   原因：因为while(begin<=end)，而且：begin=mid或begin=end，所以当mid=begin而且找不到key的时候会死循环。

正确的写法：

       public static int binary_search(int [] array, int key, int begin, int end) {
        int mid;
        while(begin<=end) {
            mid = begin + (end - begin)/2;
            //System.out.println("begin = " + begin + " mid = " + mid + " end = " + end);
            if (array[mid]==key) {
                return mid;
            }
        
            if(array[mid]>key) {
                end = mid -1;
            }
            else {
                begin = mid + 1;
            }
        }
        return 0;
    }

        int array[] = {1, 2, 3, 3, 5, 7, 8, 13, 15,19};
        int index = BinarySearch.search(array, 5, 0, 10);

执行结果：

        begin = 0 mid = 5 end = 10
begin = 0 mid = 2 end = 4
begin = 3 mid = 3 end = 4
begin = 4 mid = 4 end = 4
index = 4