zoukankan      html  css  js  c++  java
  • LeetCode第一题—— Two Sum(寻找两数,要求和为target)

    题目描述:

    Given an array of integers, return indices of the two numbers such that they add up to a specific target.

    You may assume that each input would have exactly one solution, and you may not use the same element twice.

    Example:

    Given nums = [2, 7, 11, 15], target = 9,
    
    Because nums[0] + nums[1] = 2 + 7 = 9,
    return [0, 1].

    My solution(50ms,38.5MB)

    class Solution {
        public int[] twoSum(int[] nums, int target) {
            int[] result = new int[2];
            for(int i=0;i<nums.length;i++){
                for(int j=i+1;j<nums.length;j++){
                    if(nums[i]+nums[j]==target){
                        result[0] = i;
                        result[1] = j;
                    }
                }
            }
            return result;
        }
    }

    以下是标准答案

    Approach 1: Brute Force(16ms,38.5MB)

    class Solution {
        public int[] twoSum(int[] nums, int target) {
            for (int i = 0; i < nums.length; i++) {
                for (int j = i + 1; j < nums.length; j++) {
                    if (nums[j] == target - nums[i]) {
                        return new int[] { i, j };
                    }
                }
            }
            throw new IllegalArgumentException("No two sum solution");
       }
    }

    Approach 2: Two-pass Hash Table(2ms,38.1MB)

    class Solution {
          public int[] twoSum(int[] nums, int target) {
              Map<Integer, Integer> map = new HashMap<>();
              for (int i = 0; i < nums.length; i++) {
                  map.put(nums[i], i);
              }
              for (int i = 0; i < nums.length; i++) {
                  int complement = target - nums[i];
                  if (map.containsKey(complement) && map.get(complement) != i) {
                      return new int[] { i, map.get(complement) };
                  }
              }
              throw new IllegalArgumentException("No two sum solution");
          }
    }

    Approach 3: One-pass Hash Table(2ms,38.2MB)

    class Solution {
        public int[] twoSum(int[] nums, int target) {
            Map<Integer, Integer> map = new HashMap<>();
            for (int i = 0; i < nums.length; i++) {
                int complement = target - nums[i];
                if (map.containsKey(complement)) {//判断键名是否包含complement
                    return new int[] { map.get(complement), i };
                }
                map.put(nums[i], i);
            }
            throw new IllegalArgumentException("No two sum solution");
        }
    }
  • 相关阅读:
    pip 查看软件包 可用版本并安装; pip 查看 numpy 可用版本并安装
    git submodule git 子模块管理相关操作
    git 取消文件跟踪
    C 实战练习题目1
    VC6.0设置注释快捷键
    2019-11-29-WPF-非客户区的触摸和鼠标点击响应
    2019-11-29-VisualStudio-使用新项目格式快速打出-Nuget-包
    2019-11-29-WPF-从触摸消息转触摸事件
    2019-11-29-win7-无法启动-WPF-程序-D3Dcompiler_47.dll-丢失
    2019-11-29-C#-序列类为-xml-可以使用的特性大全
  • 原文地址:https://www.cnblogs.com/mgblog/p/10653391.html
Copyright © 2011-2022 走看看