Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, where index1 m
https://leetcode.com/problems/3sum/ 套路比较常见了,最重要的是去重.还是没法一次通过. class Solution { public: vector<vector<int>> threeSum(vector<int>& a) { vector<vector<int>> ans; int n = a.size(); ) return ans; sort(a.begin(),a.end()); ; i &
#include <iostream>using namespace std; int main(){ //求两数之和 int a,b,sum; a=11; b=22; sum=a+b; cout<<"两个数a与b的和是"<<"sum="<<sum;} compare with the up program, think the output? #include <iostream>using namesp
题目: 给定一个数组,求如果排序之后,相邻两数的最大差值,要求时间复杂度为O(N),且要求不能用非基于比较的排序 public static int maxGap(int nums[]) { if (nums == null || nums.length < 2) { return 0; } int len = nums.length; int max = Integer.MIN_VALUE; int min = Integer.MAX_VALUE; for (int i = 0; i < l
B. Powers of Two You are given n integers a1, a2, ..., an. Find the number of pairs of indexes i, j (i < j) that ai + aj is a power of 2 (i. e. some integer xexists so that ai + aj = 2x). Input The first line contains the single positive integer n
Under Attack II Time Limit: 5 Seconds Memory Limit: 65536 KB Because of the sucessfully calculation in Under Attack I, Doctor is awarded with Courage Cross and promoted to lieutenant. But the war seems to end in never, now Doctor has a new order