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Refactor the articles related to searching algorithm. Add the chapter of binary search. Add the section of searching algorithm revisited. (#464)
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@ -1,3 +1,3 @@
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add_executable(binary_search binary_search.cpp)
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add_executable(hashing_search hashing_search.cpp)
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add_executable(leetcode_two_sum leetcode_two_sum.cpp)
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add_executable(linear_search linear_search.cpp)
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@ -1,59 +0,0 @@
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/**
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* File: binary_search.cpp
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* Created Time: 2022-11-25
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* Author: Krahets (krahets@163.com)
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*/
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#include "../include/include.hpp"
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/* 二分查找(双闭区间) */
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int binarySearch(vector<int> &nums, int target) {
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// 初始化双闭区间 [0, n-1] ,即 i, j 分别指向数组首元素、尾元素
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int i = 0, j = nums.size() - 1;
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// 循环,当搜索区间为空时跳出(当 i > j 时为空)
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while (i <= j) {
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int m = (i + j) / 2; // 计算中点索引 m
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if (nums[m] < target) // 此情况说明 target 在区间 [m+1, j] 中
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i = m + 1;
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else if (nums[m] > target) // 此情况说明 target 在区间 [i, m-1] 中
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j = m - 1;
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else // 找到目标元素,返回其索引
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return m;
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}
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// 未找到目标元素,返回 -1
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return -1;
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}
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/* 二分查找(左闭右开) */
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int binarySearch1(vector<int> &nums, int target) {
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// 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
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int i = 0, j = nums.size();
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// 循环,当搜索区间为空时跳出(当 i = j 时为空)
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while (i < j) {
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int m = (i + j) / 2; // 计算中点索引 m
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if (nums[m] < target) // 此情况说明 target 在区间 [m+1, j) 中
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i = m + 1;
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else if (nums[m] > target) // 此情况说明 target 在区间 [i, m) 中
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j = m;
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else // 找到目标元素,返回其索引
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return m;
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}
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// 未找到目标元素,返回 -1
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return -1;
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}
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/* Driver Code */
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int main() {
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int target = 6;
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vector<int> nums = {1, 3, 6, 8, 12, 15, 23, 67, 70, 92};
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/* 二分查找(双闭区间) */
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int index = binarySearch(nums, target);
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cout << "目标元素 6 的索引 = " << index << endl;
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/* 二分查找(左闭右开) */
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index = binarySearch1(nums, target);
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cout << "目标元素 6 的索引 = " << index << endl;
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return 0;
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}
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53
codes/cpp/chapter_searching/leetcode_two_sum.cpp
Normal file
53
codes/cpp/chapter_searching/leetcode_two_sum.cpp
Normal file
@ -0,0 +1,53 @@
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/**
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* File: leetcode_two_sum.cpp
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* Created Time: 2022-11-25
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* Author: Krahets (krahets@163.com)
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*/
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#include "../include/include.hpp"
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/* 方法一:暴力枚举 */
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vector<int> twoSumBruteForce(vector<int> &nums, int target) {
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int size = nums.size();
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// 两层循环,时间复杂度 O(n^2)
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for (int i = 0; i < size - 1; i++) {
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for (int j = i + 1; j < size; j++) {
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if (nums[i] + nums[j] == target)
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return {i, j};
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}
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}
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return {};
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}
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/* 方法二:辅助哈希表 */
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vector<int> twoSumHashTable(vector<int> &nums, int target) {
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int size = nums.size();
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// 辅助哈希表,空间复杂度 O(n)
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unordered_map<int, int> dic;
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// 单层循环,时间复杂度 O(n)
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for (int i = 0; i < size; i++) {
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if (dic.find(target - nums[i]) != dic.end()) {
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return {dic[target - nums[i]], i};
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}
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dic.emplace(nums[i], i);
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}
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return {};
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}
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int main() {
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// ======= Test Case =======
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vector<int> nums = {2, 7, 11, 15};
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int target = 9;
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// ====== Driver Code ======
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// 方法一
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vector<int> res = twoSumBruteForce(nums, target);
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cout << "方法一 res = ";
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printVector(res);
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// 方法二
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res = twoSumHashTable(nums, target);
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cout << "方法二 res = ";
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printVector(res);
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return 0;
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}
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