LeetCode算法(Python)--1、Two Sum

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].

这种蛮力方法是：

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]

另外一种比较好的做法是:

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        arr = {};//使用字典存储遍历过的num和对应下标{num:index}
        length = len(nums);//计算输入的列表长度
        for i in range(length):
            if (target - nums[i]) in arr://如果target-当前num的差在arr中，则表示已经找到答案，返回结果即可  
                return [arr[target - nums[i]], i];
            arr[nums[i]] = i;//否则，将该num及其下标存入arr中