二进制搜索不运行？

Question

我尝试在我的代码示例中使用二进制搜索算法，但它没有按我的预期运行。我不知道为什么。请帮我解释一下

var array = [1, 4, 6, 8, 9, 12, 15, 17, 19, 34, 55, 78, 80];

function binarySearch (array, numberToSearch) {
var firstIndex = 0;
var lastIndex = array.length - 1;
var currentIndex;
var currentElement;

currentIndex = (lastIndex + firstIndex) / 2 | 2;
currentElement = array[currentIndex];

while (firstIndex <= lastIndex) {
    if (numberToSearch === currentElement) {
        // found
        console.log(currentIndex);
        return currentIndex;
    } else if (numberToSearch < currentElement) {
        lastIndex = currentIndex - 1;
        currentIndex = (lastIndex + firstIndex) / 2 | 2;
        currentElement = array[currentIndex];
    } else if (numberToSearch > currentElement) {
        firstIndex = currentIndex + 1;
        currentIndex = (lastIndex + firstIndex) / 2 | 2;
        currentElement = array[currentIndex];
    }
}
 return -1;
}
binarySearch(array, 12);

我应该打印：5 但什么也没发生

Answer 1

你的代码有什么问题：

1. 应该是var currentIndex = (lastIndex + firstIndex) / 2 | 0; （不是| 2）；

2. currentIndex 和 currentElement 应该在每个 循环内的迭代。

所以，这是您的代码的更正版本：

var array = [1, 4, 6, 8, 9, 12, 15, 17, 19, 34, 55, 78, 80];

function binarySearch (array, numberToSearch) {
var firstIndex = 0;
var lastIndex = array.length - 1;

while (firstIndex <= lastIndex) {
    var currentIndex = (lastIndex + firstIndex) / 2 | 0;
    var currentElement = array[currentIndex];
    if (numberToSearch === currentElement) {
        return currentIndex;
    } else if (numberToSearch < currentElement) {
        lastIndex = currentIndex - 1;
    } else if (numberToSearch > currentElement) {
        firstIndex = currentIndex + 1;
    }
}
 return -1;
}

console.log(binarySearch(array, 99)); // -1
console.log(binarySearch(array, 12)); // 5

顺便说一句，这个var currentIndex = (lastIndex + firstIndex) / 2 | 0; 看起来很不寻常。常用方式为var currentIndex = Math.floor((lastIndex + firstIndex) / 2);

Answer 2

这是我更新的所有地方的正确答案。你非常接近！似乎您正在将二进制搜索的iterative 版本与recursive 解决方案结合起来。

CodePen Demo

var array = [1, 4, 6, 8, 9, 12, 15, 17, 19, 34, 55, 78, 80];

function binarySearch (array, numberToSearch) {
  var firstIndex = 0;
  var lastIndex = array.length - 1;
  var currentIndex;
  var currentElement;

  while (firstIndex <= lastIndex) {
    currentIndex = (lastIndex + firstIndex) / 2 | 0; //should default to zero, not Two! 
    currentElement = array[currentIndex];//These should both update every iteration so it does not infinitely loop!
    if (numberToSearch === currentElement) {
      // found
      console.log(currentIndex);
      return currentIndex;
    }else if (numberToSearch < currentElement) {
      lastIndex = currentIndex - 1; //If current is too big, move right pointer to the left
    }else if (numberToSearch > currentElement) {
      firstIndex = currentIndex + 1;//If current is too small, move left pointer to the right
    }
  }
  return -1;//When condition of while it broken and no solution has been found, return -1 to indicate it is not in the array
}
binarySearch(array, 12) //5

不进一步优化了 Math.floor() 作为 number | 0 is sometimes faster.

Answer 3

首先，这里的按位OR 有什么意义：

currentIndex = (lastIndex + firstIndex) / 2 | 2;

我认为应该是：

currentIndex = (lastIndex + firstIndex) / 2;

搜索空间更新存在基本的缺陷。 当numberToSearch < currentElement 表示middle element(当前索引处的元素) 大于要搜索的数字时，因此，正确的新边界是：

lastIndex = currentIndex - 1;

另外，当numberToSearch > currentElement 表示middle element(当前索引处的元素) 小于要搜索的数字时，因此，正确的新边界是： p>

firstIndex = currentIndex + 1;

因此，正确的代码是：

var array = [1, 4, 6, 8, 9, 12, 15, 17, 19, 34, 55, 78, 80];

function binarySearch (array, numberToSearch) {
var firstIndex = 0;
var lastIndex = array.length - 1;
var currentIndex;
var currentElement;

currentIndex = (lastIndex + firstIndex) / 2;
currentElement = array[currentIndex];

while (firstIndex <= lastIndex) {
    if (numberToSearch === currentElement) {
        // found
        console.log(currentIndex);
        return currentIndex;
    } else if (numberToSearch < currentElement) {
        lastIndex = currentIndex - 1;
        currentIndex = (lastIndex + firstIndex) / 2 | 2;
        currentElement = array[currentIndex];
    } else if (numberToSearch > currentElement) {
        firstIndex = currentIndex + 1;
        currentIndex = (lastIndex + firstIndex) / 2 | 2;
        currentElement = array[currentIndex];
    }
}
 return -1;
}
binarySearch(array, 12);