LeetCode-0001.TwoSum
LeetCode-0001.TwoSum
!! 数据结构和算法是每个程序员必须要掌握的知识，LeetCode 平台上的算法题种类和数量也足够多。从今天开始，我会按分类进行刷题，锻炼一下自己的脑袋，并且以图文并茂的形式把笔记写下来。首先从简单的数组专题开始，所有代码以及测试用例都会上传到 github上，欢迎查阅：https://github.com/CPythoner/LeetCode。
1. 描述
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Example 1:
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Input: nums = [3,2,4], target = 6
Output: [1,2]
Example 3:
Input: nums = [3,3], target = 6
Output: [0,1]
Constraints:
2 <= nums.length <= 103-109 <= nums[i] <= 109-109 <= target <= 109- Only one valid answer exists.
2. 分析及实现
假设有一个数组 a=[2, 7, 11, 15]，寻找和为 26 的两个元素的下标。
1. 暴力法
暴力法是最容易想到的解决方法，使用两个for循环分别取值相加，计算和是否为 26。此种方案的时间复杂度为 O(n2)
假设数组的 size 为 n，那么 最外层 i 的取值范围为 [0,n-1)，内层 j 的取值范围为 [i+1, n-1]。
代码如下：
vector<int> twoSum(vector<int>& nums, int target) {
size_t size_of_nums = nums.size();
int i = 0, j = 0;
for (; i < size_of_nums - 1; i++)
{
for (j = i + 1; j < size_of_nums; j++)
{
if (nums[i] + nums[j] == target)
{
return {i, j};
}
}
}
return {};
}
2. Hash 法
另一种解法是首先将数组的值都放到一个Hash Map里，数组的值作为 key 值，数组元素的下标作为 value 值；然后遍历数组去找与其对应对的另一元素是否存在 Map 中。这个解决方案的时间复杂度为 O(n)。
注意：遍历数组时 i 的取值范围为 [0, n-1)，因为前面的数组元素都和最后一个元素进行了匹配。
代码如下：
vector<int> twoSumWithHash(vector<int>& nums, int target) {
size_t size_of_nums = nums.size();
unordered_map<int, int> mymap;
for (int i = 0; i < size_of_nums; i++)
{
mymap[nums[i]] = i;
}
for (int i = 0; i < size_of_nums - 1; i++)
{
const int another = target - nums[i];
if (mymap.find(another) != mymap.end() && mymap[another] > i)
{
return {i, mymap[another]};
}
}
return {};
}
3. Hash 法改进
上述 Hash 解法需要遍历数组两次，其实可以只遍历一次，不需要一开始就建立好 Map，而是在后面遍历的过程中创建 Map。执行流程如下：
需要注意的是：最后返回两个下标要注意顺序，在 Map 中的元素下标是小于正在遍历的数组元素的下标。
代码如下：
vector<int> twoSumWithHash2(vector<int>& nums, int target) {
unordered_map<int, int> mymap;
for (int i = 0; i < nums.size(); i++)
{
const int another = target - nums[i];
if (mymap.find(another) != mymap.end())
{
return {mymap[another], i};
}
mymap[nums[i]] = i;
}
return {};
}
3. 测试
使用 Catch2 进行测试，项目配置详见 github[1]
#include "0001. Two Sum.h"
#include <catch2/catch_test_macros.hpp>
#include <iostream>
#include "../utils/utils.h"
using namespace std;
TEST_CASE("Two Sum", "[twoSum]")
{
Solution solution;
SECTION("1")
{
vector<int> expected_result{0, 1};
vector<int> nums{2, 7, 11, 15};
vector<int> result = solution.twoSum(nums, 9);
REQUIRE(CompareVectors<int>(expected_result, result) == true);
}
SECTION("2")
{
vector<int> expected_result{1, 3};
vector<int> nums{2, 7, 11, 15};
vector<int> result = solution.twoSum(nums, 22);
REQUIRE(CompareVectors<int>(expected_result, result) == true);
}
}
TEST_CASE("Two Sum Hash", "[Two Sum Hash]")
{
Solution solution;
SECTION("1")
{
vector<int> expected_result{0, 1};
vector<int> nums{2, 7, 11, 15};
vector<int> result = solution.twoSumWithHash(nums, 9);
REQUIRE(CompareVectors<int>(expected_result, result) == true);
}
SECTION("2")
{
vector<int> expected_result{1, 3};
vector<int> nums{2, 7, 11, 15};
vector<int> result = solution.twoSumWithHash(nums, 22);
REQUIRE(CompareVectors<int>(expected_result, result) == true);
}
}
TEST_CASE("Two Sum Hash2", "[Two Sum Hash2]")
{
Solution solution;
SECTION("1")
{
vector<int> expected_result{0, 1};
vector<int> nums{2, 7, 11, 15};
vector<int> result = solution.twoSumWithHash2(nums, 9);
REQUIRE(CompareVectors<int>(expected_result, result) == true);
}
SECTION("2")
{
vector<int> expected_result{1, 3};
vector<int> nums{2, 7, 11, 15};
vector<int> result = solution.twoSumWithHash2(nums, 22);
REQUIRE(CompareVectors<int>(expected_result, result) == true);
}
}
参考资料
[1]
github: https://github.com/CPythoner/LeetCode