[LeetCode Brush Questions] - Sum of Numbers topic (more topics to be added)

1 两数之和

1 两数之和
思路：hush map

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        hush=[]
        for i in range(len(nums)):
            temp=target-nums[i]
            if temp in hush:
                return [hush.index(temp),i]
            else:
                hush.append(nums[i])

Attached is a superfluous wrong solution：
The question asked to return is index ,It doesn't apply as soon as it comes up sort 的解法,
由于numsThere are repeated elements,Unsurprisingly, this method fails.

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        temp1=nums.copy()
        nums.sort()
        l,r=0,len(nums)-1
        while l<r:
            temp=nums[l]+nums[r]
            if temp==target:
                return [temp1.index(nums[l]),temp1.index(nums[r])]
            elif temp<target:
                l+=1
            else:
                r-=1

15 三数之和

15 三数之和
思路：sort use double pointer after
三数之和题解（One of them timed out）

18 四数之和

[18 四数之和](https://leetcode.cn/problems/4sum/
思路：还是sort 之后双指针

# TODO:加上代码

633 平方数之和

633 平方数之和
这个题挺简单的,Determine the search scope is range(1, int(math.sqrt(c) + 1)),
Find two numbers in this range,Make their sum of squares the target value,
Next comes the classic sum of two numbers problem.

Python 开方：math.sqrt(num)

解法1：双指针法

class Solution(object):
    def judgeSquareSum(self, c):
        if c == 0: return True
        n=int(math.sqrt(c))
        a,b=0,n
        while a<=b:
            temp=c-a**2-b**2
            if temp ==0:
                return True
            elif temp>0:
                a+=1
            else:
                b-=1
        return False

解法2：暴力计算
注：因为只要bIt is a natural number that meets the requirements of the question,So you don't have to go through it hardlist了,直接判断b是否是整数.

Determine whether a number is an integer：int(num)==num

class Solution(object):
    def judgeSquareSum(self, c):
        if c == 0: return True
        for a in range(1, int(math.sqrt(c) + 1)):
            b = c - a * a
            if int(math.sqrt(b)) ** 2 == b:
                return True
        return False

A little thought about double pointers

What is the essence of double pointer？
Will the correct answer be clipped when moving？

珠玉在前,Let's put the analysis of double pointers written by God