leetcode- 两数之和

最容易想到的

import numpy as np
a=np.random.randint(100,size=10)

n=len(a)
c=a[n-2]+a[n-1]



for i in range(n):
    for j in range(i,n):
        if a[i]+a[j]==c:
            print(i,j)

import numpy as np
a=np.random.randint(100,size=10)

n=len(a)
c=a[n-2]+a[n-1]

d={}


temp=[]
for i in range(n):
    if a[i] in d:
        print(i,d[a[i]])
        temp.append([i,d[a[i]]])
    else:
        d[c-a[i]]=i
    

print(a[temp[0][0]],'+',a[temp[0][1]],'=',c)    
#

# -*- coding: utf-8 -*-
"""
Spyder Editor

This is a temporary script file.
"""
import time
import numpy as np


'''


class Solution:
    def twoSum(self,nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        #用len()方法取得nums列表的长度
        n = len(nums)
        #x取值从0一直到n（不包括n）
        for x in range(n):
            #y取值从x+1一直到n（不包括n）
            #用x+1是减少不必要的循环,y的取值肯定是比x大
            for y in range(x+1,n):
                #假如 target-nums[x]的某个值存在于nums中
                if nums[y] == target - nums[x]:
                    #返回x和y
                    return x,y
                    break
                else:
                    continue
'''


class Solution:
    def twoSum(self,nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        #用len()方法取得nums列表长度
        n = len(nums)
        #创建一个空字典
        d = {}
        
        for x in range(n):
            a = target - nums[x]
            #字典d中存在nums[x]时
            if nums[x] in d:
                return d[nums[x]],x
            #否则往字典增加键/值对
            else:
                d[a] = x
#                print(d)
        
        #边往字典增加键/值对，边与nums[x]进行对比
 
        
                    
a=np.random.randint(10000000,size=100000)

n=len(a)
c=a[n-2]+a[n-1]


cc=Solution()
t1=time.time()
cc1=cc.twoSum(list(a),c)
print((t1-time.time())*1000,'毫秒')

print(cc1)