递归二分查找（整数）

Question

我已经编写了这个基于递归的二进制搜索算法，但我无法理解错误的原因。

    // Only a sorted array must be entered for binary search to work
    public int binarySearch(int searchFor, int[] inArray, int from, int to){
        if (to >= from){
            int mid = (to-from)/2 + from;
            
            if (inArray[mid] == searchFor){
                return inArray[mid];
            } else if (inArray[mid] < searchFor){
                binarySearch(searchFor, inArray, ++mid, to);
            } else if (inArray[mid] > searchFor){
                binarySearch(searchFor, inArray, from, ++mid);
            }
        }
        return -1;
    }

线程“主”java.lang.StackOverflowError 中的异常

at search.binarySearch(search.java:24)
at search.binarySearch(search.java:24)
at search.binarySearch(search.java:24)
at search.binarySearch(search.java:24)

Answer 1

几个问题和评论：

• 在最后一种情况下，++mid 的值作为 binarySearch 的参数是错误的。在这种情况下，该值应该比mid 少一。

• 我也会在此处投票反对使用++mid 或--mid，因为这表明mid 更改值很重要，这不是必需的。你应该在第一种情况下传递mid+1，在第二种情况下传递mid-1。

• binarySearch 返回一个 int，但是当您的代码递归调用它时，它不会对该返回值做任何事情。

  你应该返回那个值。所以：

              } else if (inArray[mid] < searchFor){
                  return binarySearch(searchFor, inArray, mid+1, to);
  //              ^^^^^^                                  ^^^^^
              } else if (inArray[mid] > searchFor){
                  return binarySearch(searchFor, inArray, from, mid-1);
  //              ^^^^^^                                        ^^^^^
              }
  

• 表达式(to-from)/2 + from 是(from+to)/2 的冗长方式...

Answer 2

除了在binarySearch的递归调用中添加return之外，还有一些逻辑缺陷：

• mid 应该递减以捕获左侧的值：

    return binarySearch(searchFor, inArray, from, --mid);

• 应该检查有效的mid、from、to 值以适合输入数组

因此，该方法应如下所示：

public static int binarySearch(int searchFor, int[] inArray, int from, int to) {
        if (to >= from && from > -1 && to <= inArray.length) {
            int mid = (to-from)/2 + from;
            
            if (mid >= inArray.length) {
                return -1;
            }
            
            // System.out.printf("from=%d to=%d mid=%d val=%d%n", from, to, mid, inArray[mid]); // debug print
            if (inArray[mid] == searchFor) {
                return inArray[mid];
            } else if (inArray[mid] < searchFor){
                return binarySearch(searchFor, inArray, ++mid, to);
            } else {
                return binarySearch(searchFor, inArray, from, --mid);
            }
        }
        return -1;
    }
}

测试：

public static int binarySearch(int searchFor, int... inArray) {
    System.out.printf("Searching for %d in %s%n", searchFor, Arrays.toString(inArray));
    return binarySearch(searchFor, inArray, 0, inArray.length);
}

System.out.println(binarySearch(10));
    System.out.println(binarySearch(10, 10));
    System.out.println(binarySearch(10, 1));
    System.out.println(binarySearch(10, 0, 1, 5, 8, 10, 21));
    System.out.println(binarySearch( 0, 0, 1, 5, 8, 10, 21));
    System.out.println(binarySearch(21, 0, 1, 5, 8, 10, 21));
    
    System.out.println(binarySearch(10, 0, 1, 3, 5, 8, 10, 21));
    System.out.println(binarySearch( 0, 0, 1, 5, 8, 10, 15, 21));
    System.out.println(binarySearch(21, 0, 1, 5, 8, 10, 16, 21));
    
    System.out.println(binarySearch(30, 0, 1, 5, 8, 10, 21));
    System.out.println(binarySearch(-1, 0, 1, 5, 8, 10, 21));
    System.out.println(binarySearch(7, 0, 1, 5, 8, 10, 21));

输出：

Searching for 10 in []
-1
Searching for 10 in [10]
10
Searching for 10 in [1]
-1
Searching for 10 in [0, 1, 5, 8, 10, 21]
10
Searching for 0 in [0, 1, 5, 8, 10, 21]
0
Searching for 21 in [0, 1, 5, 8, 10, 21]
21
Searching for 10 in [0, 1, 3, 5, 8, 10, 21]
10
Searching for 0 in [0, 1, 5, 8, 10, 15, 21]
0
Searching for 21 in [0, 1, 5, 8, 10, 16, 21]
21
Searching for 30 in [0, 1, 5, 8, 10, 21]
-1
Searching for -1 in [0, 1, 5, 8, 10, 21]
-1
Searching for 7 in [0, 1, 5, 8, 10, 21]
-1