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feat: Revised the book (#978)

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* Sync recent changes to the revised Word.

* Revised the preface chapter

* Revised the introduction chapter

* Revised the computation complexity chapter

* Revised the chapter data structure

* Revised the chapter array and linked list

* Revised the chapter stack and queue

* Revised the chapter hashing

* Revised the chapter tree

* Revised the chapter heap

* Revised the chapter graph

* Revised the chapter searching

* Reivised the sorting chapter

* Revised the divide and conquer chapter

* Revised the chapter backtacking

* Revised the DP chapter

* Revised the greedy chapter

* Revised the appendix chapter

* Revised the preface chapter doubly

* Revised the figures
This commit is contained in:
Yudong Jin
2023-12-02 06:21:34 +08:00
committed by GitHub
parent b824d149cb
commit e720aa2d24
404 changed files with 1537 additions and 1558 deletions
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6
codes/cpp/chapter_searching/binary_search.cpp
View File

@ -24,9 +24,9 @@ int binarySearch(vector<int> &nums, int target) {
return -1;
}
/* 二分查找（左闭右开） */
/* 二分查找（左闭右开区间） */
int binarySearchLCRO(vector<int> &nums, int target) {
// 初始化左闭右开 [0, n) ，即 i, j 分别指向数组首元素、尾元素+1
// 初始化左闭右开区间 [0, n) ，即 i, j 分别指向数组首元素、尾元素+1
int i = 0, j = nums.size();
// 循环，当搜索区间为空时跳出（当 i = j 时为空）
while (i < j) {
@ -51,7 +51,7 @@ int main() {
int index = binarySearch(nums, target);
cout << "目标元素 6 的索引 = " << index << endl;
/* 二分查找（左闭右开） */
/* 二分查找（左闭右开区间） */
index = binarySearchLCRO(nums, target);
cout << "目标元素 6 的索引 = " << index << endl;

6
codes/cpp/chapter_searching/two_sum.cpp
View File

@ -9,7 +9,7 @@
/* 方法一：暴力枚举 */
vector<int> twoSumBruteForce(vector<int> &nums, int target) {
int size = nums.size();
// 两层循环，时间复杂度 O(n^2)
// 两层循环，时间复杂度为 O(n^2)
for (int i = 0; i < size - 1; i++) {
for (int j = i + 1; j < size; j++) {
if (nums[i] + nums[j] == target)
@ -22,9 +22,9 @@ vector<int> twoSumBruteForce(vector<int> &nums, int target) {
/* 方法二：辅助哈希表 */
vector<int> twoSumHashTable(vector<int> &nums, int target) {
int size = nums.size();
// 辅助哈希表，空间复杂度 O(n)
// 辅助哈希表，空间复杂度为 O(n)
unordered_map<int, int> dic;
// 单层循环，时间复杂度 O(n)
// 单层循环，时间复杂度为 O(n)
for (int i = 0; i < size; i++) {
if (dic.find(target - nums[i]) != dic.end()) {
return {dic[target - nums[i]], i};