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18.4-sum.py
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#
# @lc app=leetcode id=18 lang=python3
#
# [18] 4Sum
#
# https://leetcode.com/problems/4sum/description/
#
# algorithms
# Medium (37.52%)
# Likes: 11827
# Dislikes: 1443
# Total Accepted: 1.2M
# Total Submissions: 3.2M
# Testcase Example: '[1,0,-1,0,-2,2]\n0'
#
# 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.
#
#
# Example 1:
#
#
# Input: nums = [1,0,-1,0,-2,2], target = 0
# Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]]
#
#
# Example 2:
#
#
# Input: nums = [2,2,2,2,2], target = 8
# Output: [[2,2,2,2]]
#
#
#
# Constraints:
#
#
# 1 <= nums.length <= 200
# -10^9 <= nums[i] <= 10^9
# -10^9 <= target <= 10^9
#
#
#
# @lc code=start
from typing import List
class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
nums.sort() # 1️⃣ 정렬
n = len(nums)
result = []
for i in range(n - 3): # 첫 번째 숫자 선택
if i > 0 and nums[i] == nums[i - 1]: # 중복 스킵
continue
for j in range(i + 1, n - 2): # 두 번째 숫자 선택
if j > i + 1 and nums[j] == nums[j - 1]: # 중복 스킵
continue
left, right = j + 1, n - 1 # 투 포인터 설정
while left < right:
total = nums[i] + nums[j] + nums[left] + nums[right]
if total == target:
result.append([nums[i], nums[j], nums[left], nums[right]])
# 중복 제거
while left < right and nums[left] == nums[left + 1]:
left += 1
while left < right and nums[right] == nums[right - 1]:
right -= 1
left += 1
right -= 1
elif total < target:
left += 1
else:
right -= 1
return result
# @lc code=end