难度: Easy
原题连接
内容描述
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].
思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******
算法来源于知乎文章——趣味算法思想
先定义一个Object类型的数据结构obj,它的key为target - numbers[i](比如数组第一项为2),value为索引。然后每次都看看obj[numbers[i]] 是否存在,如果存在,那我们就找到了这样的一组数据,返回当前索引以及obj[numbers[i]]。
代码:
/**
* @param {number[]} nums
* @param {number} target
* @return {number[]}
*/
var twoSum = function(nums, target) {
var obj = {};
for(var i=0; i< nums.length;i++) {
const item = nums[i];
if(obj[item] >= 0) {
return [obj[item], i]
} else {
obj[target - item] = i;
}
}
};