public class Solution { public int[] twoSum(int[] numbers, int target) { HashMa...
针对2Sum,先对数组排序,然后使用双指针匹配可行解就可以解决,虽然可以考虑使用HashMap加速搜索,但是对于本题使用HashMap的与否的时间复杂度都一样,都是O(nlog(n))。可以参考这个链接: 求和问题总结(leetcode 2Sum, 3Sum, 4Sum, K Sum),写的很清楚。 这里我只写了2Sum和3Sum的代码,注意要避免重复排序,同时避...
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...
Solution1: HashMap 重头戏在find()的实现, 类似two sum的思想:遍历所有的key,同时算出remain = sum - key, 我们的任务是查找 1. key 等于 remain时, 要组成一对pair的条件是map.get(key) >1 2. key不等于remain, 要组成一对pair的条件是,remain也在map中 code 1classTwoSum {// O(1) add, O(n...
*由于C++98不支持hashMap可以换成java的hashMap *也可以利用C++11的unordered_map,其是c++的hashMap*/#include<iostream>#include<vector>#include<map>usingnamespacestd; vector<int> twoSum(vector<int> &numbers,inttarget){ map<int,int>twoSumMap; ...
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...
https://leetcode.com/problems/two-sum-ii-input-array-is-sorted Read More Java O(n) publicstaticintuniquePairs(int[] nums,inttarget){ Set<Integer>set=newHashSet<Integer>(); Set<Integer> seen =newHashSet<Integer>();intcount =0;for(intnum : nums){if(set.contains(target-num) && !see...
{ let val1 = l1.map(|n| n.val).unwrap_or_default(); let val2 = l2.map(|n| n.val).unwrap_or_default(); let sum = val1 + val2 + carry; carry = sum / 10; cur.next = Some(Box::new(ListNode { val: sum % 10, next: None })); cur = cur.next.as_mut().unwrap()...
Divide two integers without using multiplication, division and mod operator. 思路分析 二分法.将除数不断增倍,而结果同样扩大两倍,直到除数的值大于被除数.然后再利用被除数减去除数最后增长到小于被除数的值,递归求出结果. 例如:123/4 4<123 4*2=8<123 8*2=16>123 16*2=32<123 ...
1. Two Sum 1.1 description Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and... LeetCode_#1 Two Sum 今天正式开启我的 LeetCode 刷题之旅～ 作为一只算法菜鸡,遇到题...