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157 lines
4.1 KiB
Markdown
157 lines
4.1 KiB
Markdown
---
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id: 61abc7ebf3029b56226de5b6
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title: 实现二分搜索
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challengeType: 1
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forumTopicId: 487618
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dashedName: implement-binary-search
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---
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# --description--
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二分搜索是一种 **O(log(n))** 效率的算法，用于在已排序的数组以查找元素。 它通过以下步骤运行：
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1. 查找已排序数组的中间值 `value`。 If `value == target` return `true` (The value has been found and the search is complete).
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1. 如果中间值 `value < target`，则在下一次比较中搜索数组的右半部分。
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1. 如果中间值 `value > target`，则在下一次比较中搜索数组的左半部分。
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1. If after searching the whole array the value is not present, return `false` (The array has been searched and the value is not in the array).
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如你所见，你连续将数组减半，这为你提供了 log(n) 效率。 对于这个挑战，我们希望你展示你的工作——你是如何到达目标值的……你走过的路！
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# --instructions--
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编写一个函数 `binarySearch` 来实现对数组的二分查找算法，返回在数组中查找目标所采用的路径（每个中间值比较）。
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该函数将一个排序的整数数组和一个目标值作为输入。 它返回一个数组，其中包含（按顺序）你在原始数组的每个减半处找到的中间值，直到你找到目标值。 目标值应该是返回数组的最后一个元素。 If the value is not found, return the string `Value Not Found`.
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例如，`binarySearch([1,2,3,4,5,6,7], 5)` 将返回 `[4,6,5]`。
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对于这个挑战，当减半时，你必须在做除法时使用 `Math.floor()`：`Math.floor(x/2)`。 这将提供一致的、可测试的路径。
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**注意：** 以下数组将用于测试：
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```js
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const testArray = [
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0, 1, 2, 3, 4, 5, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
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23, 49, 70
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];
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```
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# --hints--
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`binarySearch` 应该是一个函数。
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```js
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assert(typeof binarySearch == 'function');
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```
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`binarySearch(testArray, 0)` 应该返回 `[13, 5, 2, 0]`。
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```js
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assert.deepEqual(binarySearch(_testArray, 0), [13, 5, 2, 0]);
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```
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`binarySearch(testArray, 1)` 应该返回 `[13, 5, 2, 0, 1]`。
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```js
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assert.deepEqual(binarySearch(_testArray, 1), [13, 5, 2, 0, 1]);
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```
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`binarySearch(testArray, 2)` 应该返回 `[13, 5, 2]`。
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```js
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assert.deepEqual(binarySearch(_testArray, 2), [13, 5, 2]);
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```
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`binarySearch(testArray, 6)` 应该返回字符串 `Value Not Found`。
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```js
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assert.strictEqual(binarySearch(_testArray, 6), 'Value Not Found');
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```
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`binarySearch(testArray, 11)` 应该返回 `[13, 5, 10, 11]`。
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```js
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assert.deepEqual(binarySearch(_testArray, 11), [13, 5, 10, 11])
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```
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`binarySearch(testArray, 13)` 应该返回 `[13]`。
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```js
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assert.deepEqual(binarySearch(_testArray, 13), [13]);
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```
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`binarySearch(testArray, 70)` 应该返回 `[13, 19, 22, 49, 70]`.
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```js
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assert.deepEqual(binarySearch(_testArray, 70), [13, 19, 22, 49, 70]);
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```
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# --seed--
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## --after-user-code--
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```js
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const _testArray = [
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0, 1, 2, 3, 4, 5, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22,
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23, 49, 70
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];
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```
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## --seed-contents--
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```js
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function binarySearch(searchList, value) {
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let arrayPath = [];
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return arrayPath;
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}
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```
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# --solutions--
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```js
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let binarySearch = (searchList, value) => {
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let arrayPath = [];
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// set initial L - M - R
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let left = 0;
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let right = searchList.length - 1;
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let middle = Math.floor(right / 2);
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// if first comparison finds value
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if (searchList[middle] == value) {
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arrayPath.push(searchList[middle]);
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return arrayPath;
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}
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while (searchList[middle] !== value) {
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// add to output array
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arrayPath.push(searchList[middle]);
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// not found
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if (right < left) {
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return 'Value Not Found';
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}
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// value is in left or right portion of array
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// update L - M - R
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if (searchList[middle] > value) {
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right = middle - 1;
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middle = left + Math.floor((right - left) / 2);
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} else {
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left = middle + 1;
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middle = left + Math.floor((right - left) / 2);
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}
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// if found update output array and exit
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if (searchList[middle] == value) {
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arrayPath.push(searchList[middle]);
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break;
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}
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}
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return arrayPath;
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};
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```
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