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

    题目描述:

    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

    解题方案:

    定义一个结构体,用来存储原数组中的值和位置,然后对原数组按值进行排序,得到新数组。对新数组按首尾向中间遍历。下面是该题的代码:

     1 struct number_and_location{
     2     int number;
     3     int location;
     4     bool operator < (const number_and_location& lhs)const {
     5         return number < lhs.number;
     6     }
     7 };
     8 
     9 class Solution {
    10 public:
    11     vector<int> twoSum(vector<int> &numbers, int target) {
    12         int len_of_numbers = numbers.size();
    13         vector<int> index_of_result(2, 0);
    14         vector<number_and_location> change_of_source;
    15         for(int i = 0; i < len_of_numbers; ++i){
    16             change_of_source.push_back((number_and_location){numbers[i],i});
    17         }
    18         sort(change_of_source.begin(),change_of_source.end());
    19         int j = len_of_numbers - 1;
    20         int i = 0;
    21         while ( j != i){
    22             if(change_of_source[i].number + change_of_source[j].number == target){
    23                 if(change_of_source[i].location < change_of_source[j].location){
    24                     index_of_result[0] = change_of_source[i].location + 1;
    25                     index_of_result[1] = change_of_source[j].location + 1;
    26                 }else {
    27                     index_of_result[0] = change_of_source[j].location + 1;
    28                     index_of_result[1] = change_of_source[i].location + 1;
    29                 }
    30                 break;
    31             }else if(change_of_source[i].number + change_of_source[j].number < target){
    32                 ++i;
    33             }else {
    34                 --j;
    35             }
    36         }
    37         return index_of_result;
    38     }
    39 };

     修改2015-10-02

    用哈希表

    class Solution {
    public:
        vector<int> twoSum(vector<int>& nums, int target) {
            unordered_map<int, int> index;
            vector<int> result;
            for (size_t i = 0; i < nums.size(); ++i) {
                if (index.find(target - nums[i]) == index.end()) {
                    index[nums[i]] = i;
                } else {
                    result.push_back(index[target - nums[i]] + 1);
                    result.push_back(i + 1);
                    
                    return result;
                }
            }
        }
    };
    

      

  • 相关阅读:
    hausaufgabe--python 37 -- self in Class
    hausaufgabe--python 36-- Basic knowledge of Class
    hausaufgabe--python 35
    hausaufgabe--python 34
    hausaufgabe--python 33
    Retrofit2的使用简单介绍
    android中的Application作用
    第三章 Android控件架构与自定义控件详解
    第一章 Android体系与系统架构
    写在前面的话
  • 原文地址:https://www.cnblogs.com/skycore/p/4001103.html
Copyright © 2011-2022 走看看