zoukankan      html  css  js  c++  java
  • [leetcode]Two Sum

    问题描写叙述:

    Given an array of integers, 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 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

    You may assume that each input would have exactly one solution.

    Input: numbers={2, 7, 11, 15}, target=9
    Output: index1=1, index2=2

    基本思路:

    O(n2) runtime, O(1) space – Brute force:

    The brute force approach is simple. Loop through each element x and find if there is another value that equals to target – x. As finding another value requires looping through the rest of array, its runtime complexity is O(n2).

    O(n) runtime, O(n) space – Hash table:

    We could reduce the runtime complexity of looking up a value to O(1) using a hash map that maps a value to its index.

    代码:

    vector<int> twoSum(vector<int> &numbers, int target) {
            
            //consider the same num;
            vector<int> result; 
            map<int,int> num_pos;
            map<int,int>::iterator iter;
            for(int i = 0; i < numbers.size(); i++){
                int diff = target - numbers[i];
                iter = num_pos.find(diff);
                if(iter != num_pos.end()){
                    result.push_back(iter->second);
                    result.push_back(i+1);
                    break;
                }
                num_pos.insert(make_pair(numbers[i],i+1));
            }
            return result;
            
        }


  • 相关阅读:
    【日记】200617 今天开始写日记了
    近期未复现比赛汇总
    2021NSCTF RE WP
    2021国赛CISCN 初赛 部分REwriteup
    对流氓APP——一份礼物.apk的逆向分析
    RE之攻防世界 maze
    RE-攻防世界 logmein
    RE-攻防世界 T4 simple-unpack
    安天逆向工程课程 U1
    RE-攻防世界 simple-unpack
  • 原文地址:https://www.cnblogs.com/zhchoutai/p/6698569.html
Copyright © 2011-2022 走看看