ⓁⒸ ‧‧‧ 541. Reverse String II

541. Reverse String II 反轉字串 II]

❀ Origin

Problem

Given a string s 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.

Note

1. The string consists of lower English letters only.
2. Length of the given string and k will in the range [1, 10000]

Example

Input: s = "abcdefg", k = 2
Output: "bacdfeg"

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❀ 翻譯

問題

給定一個字串 s 和一個整數 k,
你必須從字串的一開始與每 2k 的位置開始倒轉前 k 個字元.
如果字串小於 k 個字元, 倒轉全部.
如果字串小於 2k 個字元但是大於 k 個字元, 則倒轉最前面的 k 個字元, 並將其他的照原樣保存.

注意

• 在字串裡只會包含小寫英文字母.
• 給定的字符和整數 k 的範圍在 [1, 10000]

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❀ Solution

JavaScript

/**
 * @param {string} s
 * @param {number} k
 * @return {string}
 */
var reverseStr = function(s, k) {
	//先組成 k 個字串為一組的陣列
	let arrResult = [];
	for (let i = 0; i < s.length; i += k) {
		arrResult.push(s.slice(i, i + k));
	}

	// 依題意 從位置 0 開始, 做之後 k 長度字串的倒轉, 其後每 2k 位置都要做一樣動作
	return arrResult
		.map(function(item, index) {
			if (index % 2 == 0) {
				item = reverseString(item);
			}
			return item;
		})
		.join('');
};

//倒轉字串 參考： https://github.com/xTobu/LeetCode/tree/master/JavaScript/%23344
var reverseString = function(s) {
	return s.split('').reverse().join('')
};

console.log(reverseStr('abcdefghijklmnopqrstuv', 3));