二进制搜索排序数组中不存在的数字，返回大于-1的负数 [英] Binary Search a non-existing number in a sorted array, return a larger negative number than -1

问题描述

搜索数为12时，为什么返回-6而不是-1?

When the search number is 12, why does it return -6 instead of -1?

int[] list = {2, 4, 7, 10, 11, 45, 50, 59, 60, 66, 69, 70, 79};
System.out.println("1. Index is " + Arrays.binarySearch(list, 11));
System.out.println("2. Index is " + Arrays.binarySearch(list, 12));

Result:
1. Index is 4
2. Index is -6

更新

现在我明白了，因为Arrays.binarySearch将返回

Update

Now I understand because Arrays.binarySearch will return

(-(insertion point) - 1)

如果数字不在数组中.IE12在插入5时返回(-(5)-1)= -6.感谢您的帮助.

if the number is not in the array. i.e 12 is at insertion of 5, so return (-(5) - 1) = -6. Thanks for the help.

推荐答案

您可以参考Javadoc:

You may refer to the Javadoc : Arrays.binarySearch(int[] a,int key)

它返回:

搜索键的
• 索引(如果包含在数组中)；
• 否则:(-(插入点)-1 ).

• index of the search key, if it is contained in the array;
• otherwise : (-(insertion point) - 1).

在这里插入点将是:

int[] list = {2, 4, 7, 10, 11, 45, 50, 59, 60, 66, 69, 70, 79};
         //  1^ 2^ 3^  4^  5^  

位置 5 ，所以(-5-1)= -6

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