2 Sum / 3 Sum / 4 Sum

核心思想就是利用sorted array来简单排除需要输出答案的唯一性，以及在2sum map方法在高阶的不适用性

然后可以通过sort来讲高难度问题转化成2sum的小问题

比较当前和目标之间的关系，然后移动指针

2 Sum 

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

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

You can return the answer in any order.

2 Sum sorted

Given an array of integers numbers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

Return the indices of the two numbers (1-indexed) as an integer array answer of size 2, where 1 <= answer[0] < answer[1] <= numbers.length.

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

Example 1:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

3 Sum

Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Notice that the solution set must not contain duplicate triplets.

3 Sum closest

Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example 1:

Input: nums = [-1,2,1,-4], target = 1
Output: 2
Explanation: The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

4 Sum

Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:

• 0 <= a, b, c, d < n
• a, b, c, and d are distinct.
• nums[a] + nums[b] + nums[c] + nums[d] == target

You may return the answer in any order.

Code example

class Solution:
    def threeSumClosest(self, nums: List[int], target: int) -> int:
        
        # Edge case
        if nums == None or len(nums) <= 2:
            return -1
        
        # Other case
        nums.sort()
        diff = float('inf')
        ret = 0
        
        # Iterate
        # -3 0 1 2
        for i in range(0, len(nums) - 2):
            t = target - nums[i]        # real target is defined as t
            j = i + 1
            k = len(nums) - 1
            
            while j < k:
                if abs(t - nums[j] - nums[k]) < diff:
                    ret = nums[i] + nums[j] + nums[k]
                    diff = abs(t - nums[j] - nums[k])
                if nums[j] + nums[k] == t:
                    return target
                elif nums[j] + nums[k] < t:
                    j = j + 1
                else:
                    k = k - 1
        
        return ret