LeetCode-1.Two Sum_if target - nums[i] is not in dict:-优快云博客

本文深入探讨了经典的两数之和算法问题，利用字典数据结构实现高效查找，避免重复元素使用，确保唯一解的存在。文章详细解释了字典在算法中的应用，包括键值对的设置和查找过程，通过实例演示了算法的具体实现。

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Description:

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].

考察的点：
1.映射关系：映射关系有列表list 和字典dict，列表通过编号（按顺序）访问各个值，字典通过名称（不按顺序）访问其值。
2.基本的字典操作：
d[k]=v 将值v关联到键k
k in d 检查字典d是否包含键为k的项。
3.难点：
表达式 k in d（其中d是一个字典）查找的是键而不是值！！！

class Solution:
    def twoSum(self, nums, target):
        seen={} #empty dict
        for i,n in enumerate(nums):  
            other=target-n 
            
            if other in seen: #'other' is a key,not a value
                return[seen[other],i]  #seen[other] returns a value,actually a index
            else:
                seen[n]=i  #将没用的放入到字典中，到了真有用的才能正确计数，否则seen[other]返回的是0,
            print(seen)
            print("*"*10)
                
nums=[2,7,11,15]
target=26
t=Solution()
t.twoSum(nums,target)

class Solution:
    def twoSum(self, nums, target):
        dict={}
        for i in range(len(nums)):
            print(i)
            if target-nums[i] not in dict:
                dict[nums[i]]=i  #nums[i] is a key ,i is a value
            else:
                return [dict[target-nums[i]],i] #now dict[target-nums[i]] 是倒数第二个，i is the last one.
            print(dict)
            
            print("*"*10)

nums=[2,7,11,15]
target=26
t=Solution()
t.twoSum(nums,target)