1. hashmap是单线程的,不安全,hashtable是多线程的安全 2. hashmap可以允许 值和键为空, hashtable不允许。 在下面的代码中,用的是hashtable,有点不太合适,可以改为hashmap。 Java代码: 1packagecom.hb.leetcode;23importjava.util.Hashtable;45importoffer.utilities.ArrayUtils;678/*9* Two Sum10*/11publ...
下面是C++的做法,就是转换为2Sum的问题, 代码如下: #include <iostream> #include <vector> #include <algorithm> using namespace std; class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { vector<vector<int>> res; sort(nums.begin(),nums.end()); for (int i = 0; i...
int[] twoSum(int[] nums, int target) { int n = nums.length; index<Integer, Integer> index = new HashMap<>(); // 构造一个哈希表:元素映射到相应的索引 for (int i = 0; i < n; i++) index.put(nums[i], i); for (int i = 0; i < n; i++) { int other = target - ...
而unordered_map需要定义hash_value函数并且重载operator==。但是很多系统内置的数据类型都自带这些,那么如果是自定义类型,那么就需要自己重载operator<或者hash_value()了。 1. 运行效率:unordered_map高,map效率较低,但是提供了稳定效率和有序的序列 2.内存占用:map内存占用略低,unordered_map较高,而且是线性成比例...
- 如果不存在,将当前数字及其索引存入hashmap5. **时间复杂度**:O(n),只需一次遍历6. **示例执行**:以nums=[2,7,11,15], target=9为例: - 遍历i=0时,num=2,diff=7(此时哈希表为空,存入2:0) - 遍历i=1时,num=7,diff=2(已存在于哈希表),返回[hashmap[2],1]=[0,1]反馈...
nums = [2,7,11,15] & target = 9 -> [0,1], 2 + 7 = 9 At each num, calculate complement, if exists in hash map then return Time: O(n) Space: O(n) */ class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { int n = nums.size(); unordered_map...
一直以来遇到底层自己无法解决的问题,都是通过在 Google、GitHub 上搜索组件、博客来进行解决。这样虽然挺快,但是也让自己成为了一个“Ctrl+C/Ctrl+V”程序员。从来不花时间思考技术的内在原理。 直到我刷了 Leetcode 第一道题目 Two Sum,接触到了 HashMap 的妙用,才激发起我去了解 HashMap 原理的兴趣。
jetcache:statIntervalMinutes:15areaInCacheName:falselocal:default:type:linkedhashmapkeyConvertor:fastjsonlimit:100remote:default:type:rediskeyConvertor:fastjsonvalueEncoder:javavalueDecoder:javapoolConfig:minIdle:5maxIdle:20maxTotal:50host:${redis.host}port:${redis.port} ...
Re-write the program to use Arraylist or Arrays and avoid using: import java.io.BufferedWriter; import java.util.HashMap; import java.util.Map; Problem description: Establish parking system using on JAVA Write a program that prompts the user to enter the maximum number...
Output: "python", "at", "includehelp" A simple method to solve the problem is by finding words that occur only once in any of the strings. For this, we will create a hashmap by which will store the words and their frequency of occurrence for both strings. And then print those words...