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[robinyoon-dev] WEEK 01 solutions #2356
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,17 @@ | ||
| /** | ||
| * @param {number[]} nums | ||
| * @return {boolean} | ||
| */ | ||
| var containsDuplicate = function(nums) { | ||
| let arrToSet = new Set(nums); | ||
|
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| const originalLength = nums.length; | ||
| const filteredLength = arrToSet.size; | ||
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| if(originalLength === filteredLength){ | ||
| return false; // 겹치는 숫자가 없는 경우 | ||
| } else { | ||
| return true; // 겹치는 숫자가 있는 경우 | ||
| } | ||
|
|
||
| }; |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,33 @@ | ||
| /** | ||
| * @param {number[]} nums | ||
| * @param {number} k | ||
| * @return {number[]} | ||
| */ | ||
| var topKFrequent = function(nums, k) { | ||
| let numsMap = new Map(); | ||
|
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| for (let i = 0; i < nums.length; i++) { | ||
| // 1. numsMap에 nums[i]가 key로 있나 확인 | ||
| // 1-1. 있으면 nums[i]와 동일한 key를 찾은 후 value에 +1 | ||
| // 1-2. 없으면 새로 생성 & value를 1로. | ||
| let hasNum = numsMap.has(nums[i]); | ||
| if (hasNum) { | ||
| numsMap.set(nums[i], numsMap.get(nums[i]) + 1); | ||
| } else { | ||
| numsMap.set(nums[i], 1); | ||
| } | ||
| } | ||
|
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| // 2. numsMap을 배열로 변경 (mapToArr) | ||
| // 3. mapToArr에서 value가 가장 놓은 순대로 sort | ||
| // 4. mapToArr를 k개 까지만 잘라낸 후 | ||
| // 5. mapToArr의 각 아이템의 0번째 값만 추출하여 새로운 배열 만들고 return | ||
| let mapToArr = Array.from(numsMap); | ||
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| mapToArr.sort((a, b) => b[1] - a[1]); | ||
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| const slicedMapToArr = mapToArr.slice(0, k); | ||
| const result = slicedMapToArr.map((item) => item[0]); | ||
|
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| return result; | ||
| }; |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,21 @@ | ||
| /** | ||
| * @param {number[]} nums | ||
| * @param {number} target | ||
| * @return {number[]} | ||
| */ | ||
| var twoSum = function (nums, target) { | ||
| for (let i = 0; i < nums.length; i++) { | ||
| let remaindedNum = target - nums[i]; | ||
| let matchedNum = nums.find((item) => remaindedNum === item); | ||
| let matchedNumIndex = nums.indexOf(matchedNum, i + 1); | ||
| // i와 matchedNumIndex 이 같은 숫자면 안 됨. | ||
| if (i === matchedNumIndex) { | ||
| continue; | ||
| } else if (matchedNumIndex === -1) { | ||
| continue; | ||
| } | ||
| else { | ||
| return [i, matchedNumIndex]; | ||
| } | ||
| } | ||
| }; |
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set가 unique값을 모아주니까 단순히 사이즈 비교로 파악이 가능하군요.
인사이트를 배워갑니다,,