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[혜준] Week1 문제 풀이 #308
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[혜준] Week1 문제 풀이 #308
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/** | ||
* @param {number[]} nums | ||
* @return {boolean} | ||
*/ | ||
var containsDuplicate = function (nums) { | ||
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const set = new Set(nums); | ||
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return nums.length !== set.size ? true : false; | ||
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}; | ||
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console.log(containsDuplicate([1, 2, 3, 1])); // true | ||
console.log(containsDuplicate([1, 2, 3, 4])); // false | ||
console.log(containsDuplicate([1, 1, 1, 3, 3, 4, 3, 2, 4, 2])); // true |
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Original file line number | Diff line number | Diff line change |
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/** | ||
* @param {number} n | ||
* @return {number} | ||
*/ | ||
var hammingWeight = function (n) { | ||
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let val = n.toString(2); | ||
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let res = 0; | ||
[...val].forEach((val) => res += parseInt(val)) | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 안녕하세요 @hyejjun 님! There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 네! 피드백 감사합니다! |
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return res; | ||
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}; | ||
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// O(LogN) | ||
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console.log(hammingWeight(11)); | ||
console.log(hammingWeight(128)); | ||
console.log(hammingWeight(2147483645)); |
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/** | ||
* @param {string} s | ||
* @return {number} | ||
*/ | ||
var countSubstrings = function (s) { | ||
let count = 0; | ||
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function checkPalindromic(left, right) { | ||
while (left >= 0 && right < s.length && s[left] === s[right]) { | ||
count++; | ||
left--; | ||
right++; | ||
} | ||
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} | ||
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for (let i = 0; i < s.length; i++) { | ||
checkPalindromic(i, i); | ||
checkPalindromic(i, i + 1); | ||
} | ||
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return count; | ||
}; | ||
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console.log(countSubstrings("abc")); | ||
console.log(countSubstrings("aaa")); | ||
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/* | ||
Time Complexity : O(n^2) | ||
Space Complexity: O(1) | ||
*/ |
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/** | ||
* @param {number[]} nums | ||
* @param {number} k | ||
* @return {number[]} | ||
*/ | ||
var topKFrequent = function (nums, k) { | ||
const count = {}; | ||
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nums.forEach((num) => { | ||
count[num] = (count[num] || 0) + 1; | ||
}); | ||
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const filtered = Object.keys(count).sort((a, b) => count[b] - count[a]); | ||
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return filtered.slice(0, k).map(Number); | ||
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}; | ||
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console.log(topKFrequent([1, 1, 1, 2, 2, 3], 2)); // [1, 2] | ||
console.log(topKFrequent([1], 1)); // [1] | ||
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/* | ||
Time Complexity : O(NLogN) | ||
Space Complexity: O(N) | ||
*/ |
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오 간결한 풀이네요!
매우 사소한 의견이지만, 삼항 연산자를 사용하지 않아도 될 것 같습니다!
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return nums.length !== set.size;
넵! 이해했습니다. 의견 감사합니다!