LeetCode：在一个整型数组中找到和为目标数字的两个元素的索

描述

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].

我的解决办法

遍历数组，求出目标数字与数组元素的差值diff，然后在数组中找到diff，返回其index

代码如下

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        temp = ''
        for index in range(len(nums)):
            diff = target - nums[index]
            for index_another in range(index+1, len(nums)):
                if nums[index_another] == diff:
                    temp = index_another
                    return [index, temp]
        return None

算法评估

Runtime: 3416 ms, faster than 28.78% of Python online submissions for Two Sum.
Memory Usage: 12.7 MB, less than 63.87% of Python online submissions for Two Sum.

参考算法

// python3
class Solution:
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dict = {}
        for i in range(len(nums)):
            if target-nums[i] not in dict:
                dict[nums[i]]=i
            else:
                return [dict[target-nums[i]],i]
                
// JavaScript
const twoSum = function(nums, target) {
    const comp = {};
    for(let i=0; i<nums.length; i++){
        if(comp[nums[i] ]>=0){
            return [ comp[nums[i] ] , i]
        }
        comp[target-nums[i]] = i
    }
};
## 参考链接
转载自：[https://leetcode.com/problems/two-sum/](https://leetcode.com/problems/two-sum/)