刷题笔记06-19 (2)

经典的Two Sum问题

题目如下

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

第一种解法 利用散列表

时间复杂度O(n)，空间复杂度O(n)
解题思路：
比较直接的解题思路，遍历数组的所有元素x，查找是否有值匹配target - x。利用散列表查找时间复杂度为O(1)
我们可以将查找匹配值的时间复杂度下降到O(n)，相应的，因为我们建立映射表，所以空间复杂度提升到了O(n)

def twoSum(nums,target):
    result = []
    dict1 = {}
    for i in range(0,len(nums)):
        dict1[nums[i]] = i # build the mapping between the nums elements and its index
    for n in range(0,len(nums)):
        the_other = target - nums[n] 
        if the_other in dict1 and dict1[the_other] != n:# make sure we don't find the same value
            result.append(n)
            result.append(dict1[the_other])
            # if we return result outside the loop, we may get duplicate index as we loop through twice the same combinations
            return result 

第二种解法 暴力算法

时间复杂度O(n**2)，空间复杂度O(n)
解题思路，没啥好说的。

def twoSum(self, nums, target):
        result = []
        for i in range(0,len(nums)):
            mama = target - nums[i]
            for j in range(0,len(nums)):
                if nums[j] == mama and j != i:
                    result.append(i)
                    result.append(j)
                    return result