AI代码解释 1// 对撞指针2// 时间复杂度: O(n)3// 空间复杂度: O(1)4class Solution{5public:6vector<int>twoSum(vector<int>&numbers,int target){7int l=0,r=numbers.size()-1;8while(l<r){9if(numbers[l]+numbers[r]==target){10int res[2]={l+1,r+1};11returnvector<int>(res,re...
"""hash= {}foriinrange(len(nums)):iftarget - nums[i]inhash:return[hash[target - nums[i]], i]hash[nums[i]] = ireturn[-1, -1] java 版本: classSolution{publicint[]twoSum(int[] nums,inttarget){if(nums ==null|| nums.length <=1) { System.out.println("input error, please c...
Two Sum II - Input array is sorted 参考资料: https://leetcode.com/problems/two-sum/ https://leetcode.com/problems/two-sum/discuss/3/Accepted-Java-O(n)-Solution https://leetcode.com/problems/two-sum/discuss/13/Accepted-C++-O(n)-Solution LeetCode All in One 题目讲解汇总(持续更新中.....
Given an array of integers, find how many pairs in the array such that their sum is bigger than a specific target number. Please return the number of pairs. Example Given numbers = [2, 7, 11, 15], target = 24. Return 1. (11 + 15 is the only pair) Challenge Do it in O(1) ...
Solve two sum problem (Javascript, Java, C#, Swift, Kotlin, Python, C++, Golang) 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 you may not use the same el...
Java Solution Since the desired class need add and get operations, HashMap is a good option for this purpose. publicclassTwoSum{privateHashMap<Integer, Integer>elements=newHashMap<Integer, Integer>();publicvoidadd(intnumber){if(elements.containsKey(number)){elements.put(number, elements.get(numbe...
Sample Solution: Java Code: importjava.util.*;publicclassMain{publicstaticvoidmain(String[]args){// Prompting the user to input two integers (a and b)System.out.println("Input two integers(a b):");// Creating a Scanner object for user inputScannerstdIn=newScanner(System.in);// Reading ...
Java programming exercises and solution: Write a Java program to find any number between 1 and n that can be expressed as the sum of two cubes in two (or more) different ways.
从栈中分别弹出栈顶数字 adder1 和 adder2,计算 adder1 和 adder2 之和,再加上进位 carry,得到当前位置的和 sum。 如果sum >= 10 ,那么进位 carry = 1...
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