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VSCode for LeetCode 测试用例模版
"use strict"; /** * * @author xgqfrms * @license MIT * @copyright xgqfrms * @created 2022-07-04 * @modified * * @description 459. Repeated Substring Pattern * @description 459. 重复的子字符串 * @difficulty Easy * @time_complexity O(n) * @space_complexity O(n) * @augments * @example * @link https://leetcode.com/problems/repeated-substring-pattern/ * @link https://leetcode-cn.com/problems/repeated-substring-pattern/ * @solutions * * @best_solutions * */ const log = console.log; /** * @param {string} s * @return {boolean} */ var repeatedSubstringPattern = function(s) { const len = s.length; if(len < 2) { return false; } // write your code here let temp = ''; for(let i = 0; i < (len / 2); i++) { temp += s[i]; const reg = new RegExp(temp, 'g'); const left = s.replace(reg, ''); if(!left.length) { return true; } } return false; } // 如果是的话,字符串一定是其子串的整数的倍数, 贪心算法 // 测试用例 test cases const testCases = [ { input: 'abab', result: true, desc: 'value equal to true', }, { input: 'aba', result: false, desc: 'value equal to false', }, { input: 'abcabcabcabc', result: true, desc: 'value equal to true', }, { input: 'a', result: false, desc: 'value equal to false', }, { input: 'abaababaab', // "abaab" result: true, desc: 'value equal to true', }, { input: 'babbabbabbabbab', // "bab" result: true, desc: 'value equal to true', }, ]; for (const [i, testCase] of testCases.entries()) { const result = repeatedSubstringPattern(testCase.input); log(`test case i result: `, result === testCase.result ? `✅ passed` : `❌ failed`, result); // log(`test case i =`, testCase); }
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VSCode for LeetCode TestCase Template All In One
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