Question

我正致力于汗学院的算法： https://www.khanacademy.org/computing/computer-science/algorithms/binary-search/p/challenge-binary-search 下面的大部分代码都是-1？这是为什么？所以二进制搜索不能有效地工作？

    var doSearch = function(array, targetValue) {
    var min = 0;
    var max = array.length - 1;
    var guess;

    while(min < max) {
        guess = Math.floor((max + min) / 2);

        if (array[guess] === targetValue) {
            return guess;
        }
        else if (array[guess] < targetValue) {
            min = guess + 1;
        }
        else {
            max = guess - 1;
        }

    }

    return -1;
};

var primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 
        41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97];

for(var i=0;i<primes.length;i++){
    var result = doSearch(primes,primes[i]);
    console.log("Found prime at index of " + primes[i] +" @ " + result);    
}

结果：

Found prime at index of 2 @ 0
Found prime at index of 3 @ -1
Found prime at index of 5 @ 2
Found prime at index of 7 @ 3
Found prime at index of 11 @ -1
Found prime at index of 13 @ 5
Found prime at index of 17 @ 6
Found prime at index of 19 @ -1
Found prime at index of 23 @ 8
Found prime at index of 29 @ -1
Found prime at index of 31 @ 10
Found prime at index of 37 @ -1
Found prime at index of 41 @ 12
Found prime at index of 43 @ 13
Found prime at index of 47 @ -1
Found prime at index of 53 @ 15
Found prime at index of 59 @ 16
Found prime at index of 61 @ -1
Found prime at index of 67 @ 18
Found prime at index of 71 @ 19
Found prime at index of 73 @ -1
Found prime at index of 79 @ 21
Found prime at index of 83 @ -1
Found prime at index of 89 @ 23
Found prime at index of 97 @ -1

我缺少什么？

Answer 1

您过早地终止循环 - min == max是一个有效的条件。

将循环更改为

while(min <= max) {
    guess = Math.floor((max + min) / 2);

    if (array[guess] === targetValue) {
        return guess;
    }
    else if (array[guess] < targetValue) {
        min = guess + 1;
    }
    else {
        max = guess - 1;
    }

}

我得到

的输出

VM153:30 Found prime at index of 2 @ 0
VM153:30 Found prime at index of 3 @ 1
VM153:30 Found prime at index of 5 @ 2
VM153:30 Found prime at index of 7 @ 3
VM153:30 Found prime at index of 11 @ 4
VM153:30 Found prime at index of 13 @ 5
VM153:30 Found prime at index of 17 @ 6
VM153:30 Found prime at index of 19 @ 7
VM153:30 Found prime at index of 23 @ 8
VM153:30 Found prime at index of 29 @ 9
VM153:30 Found prime at index of 31 @ 10
VM153:30 Found prime at index of 37 @ 11
VM153:30 Found prime at index of 41 @ 12
VM153:30 Found prime at index of 43 @ 13
VM153:30 Found prime at index of 47 @ 14
VM153:30 Found prime at index of 53 @ 15
VM153:30 Found prime at index of 59 @ 16
VM153:30 Found prime at index of 61 @ 17
VM153:30 Found prime at index of 67 @ 18
VM153:30 Found prime at index of 71 @ 19
VM153:30 Found prime at index of 73 @ 20
VM153:30 Found prime at index of 79 @ 21
VM153:30 Found prime at index of 83 @ 22
VM153:30 Found prime at index of 89 @ 23
VM153:30 Found prime at index of 97 @ 24

Answer 2

假设您的列表只有2个素数，并且您搜索更大的素数。你的循环将针对较小的失败进行测试，并将min设置为与max相同，因此循环将在检查该值之前终止。

你的循环警卫应该是min <= max。

汗学院算法：二元搜索解决方案

2 个答案: