xinyin025/hello-algo
1
0
Fork 0
mirror of https://github.com/krahets/hello-algo.git synced 2026-08-12 20:00:58 +00:00
Code Issues Packages Projects Releases Wiki Activity

build

Browse Source
This commit is contained in:
krahets
2024-05-01 07:30:10 +08:00
parent 583d338530
commit a08cd961b3
64 changed files with 227 additions and 227 deletions
Download Patch File Download Diff File Expand all files Collapse all files
en/docs/chapter_searching/binary_search.md
+3 -3
View File
@@ -8,13 +8,13 @@ comments: true
!!! question
Given an array `nums` of length $n$, with elements arranged in ascending order and non-repeating. Please find and return the index of element `target` in this array. If the array does not contain the element, return $-1$. An example is shown below.
Given an array `nums` of length $n$, with elements arranged in ascending order and non-repeating. Please find and return the index of element `target` in this array. If the array does not contain the element, return $-1$. An example is shown in Figure 10-1.
![Binary search example data](binary_search.assets/binary_search_example.png){ class="animation-figure" }
<p align="center"> Figure 10-1 &nbsp; Binary search example data </p>
As shown in the Figure 10-2 , we first initialize pointers $i = 0$ and $j = n - 1$, pointing to the first and last elements of the array, representing the search interval $[0, n - 1]$. Please note that square brackets indicate a closed interval, which includes the boundary values themselves.
As shown in Figure 10-2, we first initialize pointers $i = 0$ and $j = n - 1$, pointing to the first and last elements of the array, representing the search interval $[0, n - 1]$. Please note that square brackets indicate a closed interval, which includes the boundary values themselves.
Next, perform the following two steps in a loop.
@@ -739,7 +739,7 @@ We can implement a binary search algorithm with the same functionality based on
<div style="height: 549px; width: 100%;"><iframe class="pythontutor-iframe" src="https://pythontutor.com/iframe-embed.html#code=def%20binary_search_lcro%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%20%5B0,%20n%29%20%EF%BC%8C%E5%8D%B3%20i,%20j%20%E5%88%86%E5%88%AB%E6%8C%87%E5%90%91%E6%95%B0%E7%BB%84%E9%A6%96%E5%85%83%E7%B4%A0%E3%80%81%E5%B0%BE%E5%85%83%E7%B4%A0%2B1%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%EF%BC%8C%E5%BD%93%E6%90%9C%E7%B4%A2%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%E6%97%B6%E8%B7%B3%E5%87%BA%EF%BC%88%E5%BD%93%20i%20%3D%20j%20%E6%97%B6%E4%B8%BA%E7%A9%BA%EF%BC%89%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20m%20%20%23%20%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%E5%85%B6%E7%B4%A2%E5%BC%95%0A%20%20%20%20return%20-1%20%20%23%20%E6%9C%AA%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%0A%20%20%20%20index%20%3D%20binary_search_lcro%28nums,%20target%29%0A%20%20%20%20print%28%22%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%206%20%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%22,%20index%29&codeDivHeight=472&codeDivWidth=350&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false"> </iframe></div>
<div style="margin-top: 5px;"><a href="https://pythontutor.com/iframe-embed.html#code=def%20binary_search_lcro%28nums%3A%20list%5Bint%5D,%20target%3A%20int%29%20-%3E%20int%3A%0A%20%20%20%20%22%22%22%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%22%22%22%0A%20%20%20%20%23%20%E5%88%9D%E5%A7%8B%E5%8C%96%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%20%5B0,%20n%29%20%EF%BC%8C%E5%8D%B3%20i,%20j%20%E5%88%86%E5%88%AB%E6%8C%87%E5%90%91%E6%95%B0%E7%BB%84%E9%A6%96%E5%85%83%E7%B4%A0%E3%80%81%E5%B0%BE%E5%85%83%E7%B4%A0%2B1%0A%20%20%20%20i,%20j%20%3D%200,%20len%28nums%29%0A%20%20%20%20%23%20%E5%BE%AA%E7%8E%AF%EF%BC%8C%E5%BD%93%E6%90%9C%E7%B4%A2%E5%8C%BA%E9%97%B4%E4%B8%BA%E7%A9%BA%E6%97%B6%E8%B7%B3%E5%87%BA%EF%BC%88%E5%BD%93%20i%20%3D%20j%20%E6%97%B6%E4%B8%BA%E7%A9%BA%EF%BC%89%0A%20%20%20%20while%20i%20%3C%20j%3A%0A%20%20%20%20%20%20%20%20m%20%3D%20%28i%20%2B%20j%29%20//%202%20%20%23%20%E8%AE%A1%E7%AE%97%E4%B8%AD%E7%82%B9%E7%B4%A2%E5%BC%95%20m%0A%20%20%20%20%20%20%20%20if%20nums%5Bm%5D%20%3C%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20i%20%3D%20m%20%2B%201%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bm%2B1,%20j%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20elif%20nums%5Bm%5D%20%3E%20target%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20j%20%3D%20m%20%20%23%20%E6%AD%A4%E6%83%85%E5%86%B5%E8%AF%B4%E6%98%8E%20target%20%E5%9C%A8%E5%8C%BA%E9%97%B4%20%5Bi,%20m%29%20%E4%B8%AD%0A%20%20%20%20%20%20%20%20else%3A%0A%20%20%20%20%20%20%20%20%20%20%20%20return%20m%20%20%23%20%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%E5%85%B6%E7%B4%A2%E5%BC%95%0A%20%20%20%20return%20-1%20%20%23%20%E6%9C%AA%E6%89%BE%E5%88%B0%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%EF%BC%8C%E8%BF%94%E5%9B%9E%20-1%0A%0A%0A%22%22%22Driver%20Code%22%22%22%0Aif%20__name__%20%3D%3D%20%22__main__%22%3A%0A%20%20%20%20target%20%3D%206%0A%20%20%20%20nums%20%3D%20%5B1,%203,%206,%208,%2012,%2015,%2023,%2026,%2031,%2035%5D%0A%0A%20%20%20%20%23%20%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%88%E5%B7%A6%E9%97%AD%E5%8F%B3%E5%BC%80%E5%8C%BA%E9%97%B4%EF%BC%89%0A%20%20%20%20index%20%3D%20binary_search_lcro%28nums,%20target%29%0A%20%20%20%20print%28%22%E7%9B%AE%E6%A0%87%E5%85%83%E7%B4%A0%206%20%E7%9A%84%E7%B4%A2%E5%BC%95%20%3D%20%22,%20index%29&codeDivHeight=800&codeDivWidth=600&cumulative=false&curInstr=5&heapPrimitives=nevernest&origin=opt-frontend.js&py=311&rawInputLstJSON=%5B%5D&textReferences=false" target="_blank" rel="noopener noreferrer">Full Screen ></a></div>
As shown in the Figure 10-3 , in the two types of interval representations, the initialization of the binary search algorithm, the loop condition, and the narrowing interval operation are different.
As shown in Figure 10-3, in the two types of interval representations, the initialization of the binary search algorithm, the loop condition, and the narrowing interval operation are different.
Since both boundaries in the "closed interval" representation are defined as closed, the operations to narrow the interval through pointers $i$ and $j$ are also symmetrical. This makes it less prone to errors, **therefore, it is generally recommended to use the "closed interval" approach**.
en/docs/chapter_searching/binary_search_edge.md
+2 -2
View File
@@ -245,7 +245,7 @@ Below we introduce two more cunning methods.
In fact, we can use the function for finding the leftmost element to find the rightmost element, specifically by **transforming the search for the rightmost `target` into a search for the leftmost `target + 1`**.
As shown in the Figure 10-7 , after the search is completed, the pointer $i$ points to the leftmost `target + 1` (if it exists), while $j$ points to the rightmost `target`, **thus returning $j$ is sufficient**.
As shown in Figure 10-7, after the search is completed, the pointer $i$ points to the leftmost `target + 1` (if it exists), while $j$ points to the rightmost `target`, **thus returning $j$ is sufficient**.
![Transforming the search for the right boundary into the search for the left boundary](binary_search_edge.assets/binary_search_right_edge_by_left_edge.png){ class="animation-figure" }
@@ -500,7 +500,7 @@ Please note, the insertion point returned is $i$, therefore, it should be subtra
We know that when the array does not contain `target`, $i$ and $j$ will eventually point to the first element greater and smaller than `target` respectively.
Thus, as shown in the Figure 10-8 , we can construct an element that does not exist in the array, to search for the left and right boundaries.
Thus, as shown in Figure 10-8, we can construct an element that does not exist in the array, to search for the left and right boundaries.
- To find the leftmost `target`: it can be transformed into searching for `target - 0.5`, and return the pointer $i$.
- To find the rightmost `target`: it can be transformed into searching for `target + 0.5`, and return the pointer $j$.
en/docs/chapter_searching/binary_search_insertion.md
+3 -3
View File
@@ -10,7 +10,7 @@ Binary search is not only used to search for target elements but also to solve m
!!! question
Given an ordered array `nums` of length $n$ and an element `target`, where the array has no duplicate elements. Now insert `target` into the array `nums` while maintaining its order. If the element `target` already exists in the array, insert it to its left side. Please return the index of `target` in the array after insertion. See the example shown in the Figure 10-4 .
Given an ordered array `nums` of length $n$ and an element `target`, where the array has no duplicate elements. Now insert `target` into the array `nums` while maintaining its order. If the element `target` already exists in the array, insert it to its left side. Please return the index of `target` in the array after insertion. See the example shown in Figure 10-4.
![Example data for binary search insertion point](binary_search_insertion.assets/binary_search_insertion_example.png){ class="animation-figure" }
@@ -334,7 +334,7 @@ Therefore, at the end of the binary, it is certain that: $i$ points to the first
Suppose there are multiple `target`s in the array, ordinary binary search can only return the index of one of the `target`s, **and it cannot determine how many `target`s are to the left and right of that element**.
The task requires inserting the target element to the very left, **so we need to find the index of the leftmost `target` in the array**. Initially consider implementing this through the steps shown in the Figure 10-5 .
The task requires inserting the target element to the very left, **so we need to find the index of the leftmost `target` in the array**. Initially consider implementing this through the steps shown in Figure 10-5.
1. Perform a binary search, get an arbitrary index of `target`, denoted as $k$.
2. Start from index $k$, and perform a linear search to the left until the leftmost `target` is found and return.
@@ -345,7 +345,7 @@ The task requires inserting the target element to the very left, **so we need to
Although this method is feasible, it includes linear search, so its time complexity is $O(n)$. This method is inefficient when the array contains many duplicate `target`s.
Now consider extending the binary search code. As shown in the Figure 10-6 , the overall process remains the same, each round first calculates the midpoint index $m$, then judges the size relationship between `target` and `nums[m]`, divided into the following cases.
Now consider extending the binary search code. As shown in Figure 10-6, the overall process remains the same, each round first calculates the midpoint index $m$, then judges the size relationship between `target` and `nums[m]`, divided into the following cases.
- When `nums[m] < target` or `nums[m] > target`, it means `target` has not been found yet, thus use the normal binary search interval reduction operation, **thus making pointers $i$ and $j$ approach `target`**.
- When `nums[m] == target`, it indicates that the elements less than `target` are in the interval $[i, m - 1]$, therefore use $j = m - 1$ to narrow the interval, **thus making pointer $j$ approach elements less than `target`**.
en/docs/chapter_searching/replace_linear_by_hashing.md
+1 -1
View File
@@ -12,7 +12,7 @@ In algorithm problems, **we often reduce the time complexity of algorithms by re
## 10.4.1 &nbsp; Linear search: trading time for space
Consider traversing all possible combinations directly. As shown in the Figure 10-9 , we initiate a two-layer loop, and in each round, we determine whether the sum of the two integers equals `target`. If so, we return their indices.
Consider traversing all possible combinations directly. As shown in Figure 10-9, we initiate a two-layer loop, and in each round, we determine whether the sum of the two integers equals `target`. If so, we return their indices.
![Linear search solution for two-sum problem](replace_linear_by_hashing.assets/two_sum_brute_force.png){ class="animation-figure" }
en/docs/chapter_searching/searching_algorithm_revisited.md
+1 -1
View File
@@ -42,7 +42,7 @@ However, **using these algorithms often requires data preprocessing**. For examp
## 10.5.3 &nbsp; Choosing a search method
Given a set of data of size $n$, we can use linear search, binary search, tree search, hash search, and other methods to search for the target element from it. The working principles of these methods are shown in the following figure.
Given a set of data of size $n$, we can use linear search, binary search, tree search, hash search, and other methods to search for the target element from it. The working principles of these methods are shown in Figure 10-11.
![Various search strategies](searching_algorithm_revisited.assets/searching_algorithms.png){ class="animation-figure" }