usingnamespace std; vector<int>twoSum(vector<int>& nums, int target) { std:map<int, int> map; vector<int> ret; for(int i = 0; i < nums.size(); i ++) { int a = nums[i]; int b = target - a; std::map<int,int>::iterator it =map.find(b); if(it != map.end()) ...
017 two sum3-data structure design: originally, we solved this problem by hashmap, but it actually can be done using 2 pointers. this problem is really easy actually. just pay attention to details. 015 three sum: we have to do some kind of presort. and we will use three pointers 016 ...
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...
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 from the hashmap whose occurrence count i...
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...
Then, for each current number, we are looking for the difference between it and the target number. And return the indice immediately as sooon as we find it appearing in the hash table. See also:The Two Sum Algorithm using HashMap in C++/Java ...