Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them. If there are less than 2k but greater than or equal to k characters, then reverse the first k characters and left the other as original. 給定一個字符串和一個整數 k，你需要對從字符串開頭算起的每個 2k 個字符的前k個字符進行反轉。如果剩余少于 k 個字符，則將剩余的所有全部反轉。如果有小于 2k 但大于或等于 k 個字符，則反轉前 k 個字符，并將剩余的字符保持原樣。 Example: ``` Input: s = "abcdefg", k = 2 Output: "bacdfeg" ``` Restrictions: * The string consists of lower English letters only. * Length of the given string and k will in the range [1, 10000] ``` 思路：先將字符串分片（每2k為一組），然后對每一組進行長度判斷，長度小于k的和長度大于k* /** * @param {string} s * @param {number} k * @return {string} */ var reverseStr = function(s, k) { if(k <= 1){ return s; } var Arr = []; for(var i = 0 ; i < s.length; i += 2*k ){ var subStr = s.slice(i,i+2*k); var arr = subStr.split(''); if(arr.length <= k){ var str = arr.reverse().join(''); }else{ var subArr = arr.slice(0,k); subArr.reverse(); var str = subArr.concat(arr.slice(k,2*k)).join(''); } Arr[i] = str; } var result = Arr.join(''); return result; }; ```