Leetcode sum of two numbers

Ideas and algorithms

The easiest way to think of is to enumerate every number in an array x, Look for the existence of target - x.

When we look for target - x when , It needs to be noted that each is located in x All the previous elements have been associated with x Matched , So there's no need to match again . And each element cannot be used twice , So we just need to x Look for... In the following elements target - x.

#
# @lc app=leetcode.cn id=1 lang=python
#
# [1]  Sum of two numbers 
#

# @lc code=start
class Solution(object):
    def twoSum(self, nums, target):
        """ :type nums: List[int] :type target: int :rtype: List[int] """
        n = len(nums)
        for i in range(n):
            for j in range(i+1,n):
                if nums[i]+nums[j] == target:
                    return [i,j]
        return []

S = Solution()
print(S.twoSum([3,2,4],6))
# @lc code=end