二进制搜索字符串数组

Question

我想知道为什么我的二分查找返回的值与线性搜索不同。有人可以向我解释我做错了什么吗？我应该回来一些不同的东西吗？

public class BinarySearch extends SearchAlgorithm
{
    public int search (String [] words, String wordToFind) throws ItemNotFoundException {
    int lowIndex = 0;
    int highIndex = words.length - 1;
    while (lowIndex <= highIndex) {
        int midIndex = (lowIndex + highIndex) / 2;
        if ((words[midIndex]).equals(wordToFind)) {
            return midIndex;
        }
        if (wordToFind.compareTo(words[midIndex]) > 0) { //wordToFind > midIndex
            lowIndex = midIndex + 1;
        }
        if (wordToFind.compareTo(words[midIndex]) < 0) { //wordToFind < midIndex
            highIndex = midIndex - 1;
        }
        lowIndex++;
    }
    return -1;
}

}

这是它返回的内容。第一组是线性搜索，第二组是二进制。

DISCIPLINES found at index: 11780 taking 0 comparisons.
TRANSURANIUM found at index: 43920 taking 0 comparisons.
HEURISTICALLY found at index: 18385 taking 0 comparisons.
FOO found at index: -1 taking 0 comparisons.

DISCIPLINES found at index: 11780 taking 0 comparisons.
TRANSURANIUM found at index: 43920 taking 0 comparisons.
HEURISTICALLY found at index: -1 taking 0 comparisons.
FOO found at index: -1 taking 0 comparisons.

Answer 1

尝试将您的逻辑更改为：

int midIndex = (lowIndex + highIndex) / 2;
while (lowIndex <= highIndex) {    
    if ((words[midIndex]).equals(wordToFind)) { //use equals
        return midIndex;
    }
    if (wordToFind.compareTo(words[midIndex]) > 0) { //wordToFind > midIndex
        lowIndex = midIndex + 1;
    }
    if (wordToFind.compareTo(words[midIndex]) < 0) { //wordToFind < midIndex
        highIndex = midIndex - 1;
    }
// getting rid of infinite loop when the value is not in the list
    if (lowIndex==highIndex && !wordToFind.compareTo(words[midIndex]) {
        return -1;
    }
    midIndex = (lowIndex + highIndex) / 2; // notice removing lowindex++
}

lowIndex++内的while不是必需的，因为那是更新lowIndex而不管BS算法是否根据与midIndex值的比较更新索引。