Leetcode 1

題目描述：

給定一個整數陣列和一個目標值，找出陣列中和為目標值的兩個數。

你可以假設每個輸入只對應一種答案，且同樣的元素不能被重複利用。

示例:

給定 nums = [2, 7, 11, 15], target = 9

因為 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

解法一：

'''
蠻力法：兩層迴圈，暴力求解--7560ms
'''
class Solution0:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        result = []
        for i in range(len(nums)):
            for j in range(i+1,len(nums)):
                if nums[i] + nums[j] == target:
                    result.append(i)
                    result.append(j)
                    return result
 

 解法二：

'''
改進1：單層迴圈--1360ms
'''
class Solution1:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        result = []
        for i in range(len(nums)):
            second_num = target-nums[i]
            if second_num in nums:
                j = nums.index(second_num)
                if i != j:
                    result.append(i)
                    result.append(j)
                    return result
 

解法三：

'''
改進2：利用字典--48ms
'''
class Solution2:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        result = []
        index_dict = {nums[i]:i for i in range(len(nums))}
        for i in range(len(nums)):
            temp =target-nums[i]
            j = index_dict.get(temp)
            if temp in index_dict and i != j :
                result.append(i)
                result.append(j)
                return result
 

附：

# 發現先定義列表 result
result = []
# 然後append 之後再返回的效率比直接返回高 return [i,j]
     result.append(i)
     result.append(j)
     return result