leetcode 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].

solution1:

时间复杂度O(n^2)，空间复杂度O(n)

var twoSum = function(nums, target) {
    if(nums.length <=1)
        return;
    for(let i=0;i<nums.length;i++) {
        for(let j=i+1;j<nums.length;j++) {
            if(nums[i] + nums[j] === target) 
                return [i,j];
        }
    }
};

solution2:

时间复杂度O(n)，空间复杂度O(n)

var twoSum = function(nums, target) {
    if(nums.length <=1)
        return;
    var hash = {}
    for(let i=0;i<nums.length;i++) {
        j = target - nums[i];
        if(hash[j] == undefined) {               
            hash[nums[i]] = i;
        } else {
            return [hash[j],i];
        }
    }
};