leetcode1-两数字和

1. 两数之和

给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。

你可以假设每种输入只会对应一个答案。但是，数组中同一个元素不能使用两遍。

示例:

给定 nums = [2, 7, 11, 15], target = 9

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

解答：

1st:
两个循环 + 判断

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range(len(nums)):
            for j in range(len(nums)-i):
                if nums[i]+nums[j]==target:
                    return [i,j]

执行正确

 

 但提交结果错误：

改正：

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

2st：

class Solution(object):
    def twoSum(self,nums, target):
        for i in range(1, len(nums)):
            temp = nums[:i]
            num1 = target - nums[i]
            if num1 in temp:
                j = temp.index(num1)
                return [j,i]

 

 

 

3st、字典（哈希表）

class Solution(object):
    def twoSum(self, nums, target):

        hashmap = {}
        for index1, num1 in enumerate(nums):
            hashmap[num1] = index1
            num2 = target - num1
            if num2 in hashmap:
                return [hashmap[num2], index1]

 

 

 

 改正：

class Solution(object):
    def twoSum(self, nums, target):
        hashmap = {}
        for index1, num1 in enumerate(nums):
           
            num2 = target - num1
            if num2 in hashmap:
                return [hashmap[num2], index1]
            hashmap[num1] = index1 #-------