문제 링크 : https://leetcode.com/problems/two-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.
Example 1:
Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Output: Because nums[0] + nums[1] == 9, we return [0, 1].
Example 2:
Input: nums = [3,2,4], target = 6
Output: [1,2]
Example 3:
Input: nums = [3,3], target = 6
Output: [0,1]
Constraints:
2 <= nums.length <= 104
109 <= nums[i] <= 109
109 <= target <= 109
풀이
nums = [1,2,3,4,5]
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
print(i, j)
print(len(nums))
print(range(len(nums)))
print("*************************************")
print("*************************************")
for i in range(len(nums) - 1):
for j in range(i + 1, len(nums)):
print(i, j)
print(len(nums) -1)
print(range(len(nums) -1))
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
print(i, j)
print("*************************************")
print("*************************************")
for i in range(len(nums)):
for j in range(len(nums)): # 각각 1번씩만 매칭이 되면 되니까, + 1를 하는게 좋겠다.
print(i, j)
print("*************************************")
print("*************************************")
class Solution:
def twoSum(self, nums, target):
for i in range(len(nums)):
for j in range(i+1, len(nums)):
if target == nums[i] + nums[j]:
return [i, j]
solu = Solution()
print(solu.twoSum([1,2,3,1,5], 4))
# 출력
0 1
0 2
0 3
0 4
1 2
1 3
1 4
2 3
2 4
3 4
5
range(0, 5)
*************************************
*************************************
0 1
0 2
0 3
0 4
1 2
1 3
1 4
2 3
2 4
3 4
4
range(0, 4)
0 1
0 2
0 3
0 4
1 2
1 3
1 4
2 3
2 4
3 4
*************************************
*************************************
0 0
0 1
0 2
0 3
0 4
1 0
1 1
1 2
1 3
1 4
2 0
2 1
2 2
2 3
2 4
3 0
3 1
3 2
3 3
3 4
4 0
4 1
4 2
4 3
4 4
*************************************
*************************************
[0, 2]