zoukankan      html  css  js  c++  java
  • Two Sum

    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

    //辅助数据结构,保存原数组的值和索引
    struct node{
        int val;
        int pos;
    };
    bool cmp(const node &a,const node &b){
         return a.val<b.val;
        
    }
    class Solution {
    public:
        vector<int> twoSum(vector<int> &numbers, int target) {
            int i=0,j=numbers.size();
            node *nodeArray=new node[j+1];//注意此处的初始声明
            for(;i<j;i++){
                nodeArray[i].val=numbers[i];
                nodeArray[i].pos=i+1;
            }
            sort(nodeArray,nodeArray+j,cmp);//使用库函数sort,注意参数
             i=0;
             j=numbers.size()-1;
            while(i<j){
                int sum=nodeArray[i].val+nodeArray[j].val;
                if(sum==target){
                    vector<int> result;
                    
                    if(nodeArray[i].pos>nodeArray[j].pos){
                        result.push_back(nodeArray[j].pos);
                        result.push_back(nodeArray[i].pos);
                    }else{
                        result.push_back(nodeArray[i].pos);
                        result.push_back(nodeArray[j].pos);
                    }
                    return result;
                }else if(sum>target)
                        j--;
                        else
                        i++;
            }
            
        }
    };
  • 相关阅读:
    fetch
    创建Vue实例传入的option
    Text and Binary modes
    daemon_int
    http 协议 c++代码 获取网页
    asp.net mvc 5 初体验
    win32 音视频相关 api
    setuid和seteuid
    用0x077CB531计算末尾0的个数
    webservice gsoap 小记
  • 原文地址:https://www.cnblogs.com/qiaomu/p/4396841.html
Copyright © 2011-2022 走看看