51 lines
1.7 KiB
Markdown
51 lines
1.7 KiB
Markdown
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## Problem
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https://leetcode-cn.com/problems/two-sum
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## Problem Description
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```
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Given an array of integers, return indices of the two numbers such that they add up to a specific target.
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You may assume that each input would have exactly one solution, and you may not use the same element twice.
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Example:
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Given nums = [2, 7, 11, 15], target = 9,
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Because nums[0] + nums[1] = 2 + 7 = 9,
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return [0, 1].
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```
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## Solution
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The easiest solution to come up with is Brute Force. We could write two for-loops to traverse every element, and find the target numbers that meet the requirement. However, the time complexity of this solution is O(N^2), while the space complexity is O(1). Apparently, we need to find a way to optimize this solution since the time complexity is too high. What we could do is to record the numbers we have traversed and the relevant index with a Map. Whenever we meet a new number during traversal, we go back to the Map and check whether the `diff` between this number and the target number appeared before. If it did, the problem has been solved and there's no need to continue.
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## Key Points
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- Find the difference instead of the sum
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- Connect every number with its index through the help of Map
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- Less time by more space. Reduce the time complexity from O(N) to O(1)
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## Code
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- Support Language: JS
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```js
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/**
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* @param {number[]} nums
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* @param {number} target
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* @return {number[]}
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*/
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const twoSum = function (nums, target) {
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const map = new Map();
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for (let i = 0; i < nums.length; i++) {
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const diff = target - nums[i];
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if (map.has(diff)) {
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return [map.get(diff), i];
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}
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map.set(nums[i], i);
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}
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}
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```
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***Complexity Anlysis***
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- *Time Complexity*: O(N)
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- *Space Complexity*：O(N)
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