zoukankan      html  css  js  c++  java
  • 167. Two Sum II

    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 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


    从i开始枚举,然后从i+1至.size()-1二分查找,好笨的方法=。=

    #include <iostream>
    #include <vector>
    #include <set>
    #include <algorithm>
    #include <cmath>
    using namespace std;
    
    class Solution {
    public:
        vector<int> twoSum(vector<int>& numbers, int target) {
            vector<int>vec;
            for (vector<int>::size_type i = 0; i < numbers.size(); ++i)
            {
            	left = i + 1;
            	right = numbers.size() - 1;
            	int result = BinarySearch(numbers, target - numbers[i]);
            	if (numbers[i] + numbers[result] == target)
            	{
            		//cout << numbers[i] << " " << numbers[result] << endl;
            		vec.push_back(i + 1);
            		vec.push_back(result + 1);
            		break;
            	}
            }
           
       		return vec;
        }
    private:
    	int left;
    	int right;
    	int mid;
    	#define max(a,b) (((a) > (b)) ? (a) : (b))
    	#define min(a,b) (((a) > (b)) ? (b) : (a))
    	int BinarySearch(vector<int>& numbers, int target)
    	{
    		while(left <= right)
            {
            	mid = (left + right) / 2;
            	if(numbers[mid] > target)
            	{
            		right = mid - 1;
            	}
            	else if(numbers[mid] < target)
            	{
            		left = mid + 1;
            	}
            	else
            	{
            		return mid;
            	}
            }
            return right;
    	}
    };
    
    int main()
    {
       	Solution s;
       	vector<int>vec{1,2,3,4,4,9,56,90};
       	vector<int>v = s.twoSum(vec, 8);
       	for (vector<int>::size_type i = 0; i < v.size(); ++i)
       	{
       		cout << v[i] << endl;
       	}
    	return 0;
    }

    15 / 15 test cases passed.
    Status: 

    Accepted

    Runtime: 13 ms
    Submitted: 2 minutes ago

    Keep it simple!
    作者:N3verL4nd
    知识共享,欢迎转载。
  • 相关阅读:
    Nginx-->基础-->理论-->001:Nginx基本介绍
    Nginx-->基础-->理论-->nginx进程模型
    Nginx-->基础-->排错-->nginx错误总结
    Nginx-->基础-->安装-->001:安装总结
    网络-->监控-->单位换算
    网络-->监控-->OID-->BGP
    在腾讯云centOs系统上安装nginx
    react-native入门学习( 一 )
    对javascript变量提升跟函数提升的理解
    chrome浏览器好用的一些插件
  • 原文地址:https://www.cnblogs.com/lgh1992314/p/6616329.html
Copyright © 2011-2022 走看看