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[LeetCode By Go 59]541. Reverse

[LeetCode By Go 59]541. Reverse

作者: miltonsun | 来源:发表于2017-08-25 16:45 被阅读3次

题目

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.
Example:

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

Restrictions:

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

解题思路

找到需要翻转的子串翻转即可

代码

func reverseRune(r []rune, start, end int) {
    for ; start < end;  {
        r[start], r[end] = r[end], r[start]
        start++
        end--
    }
}

func reverseStr(s string, k int) string {
    sRune := []rune(s)
    len1 := len(s)
    for i := 0; i < len1; i += k * 2 {
        end := i + k - 1
        if end > len1 - 1{
            end = len1 - 1
        }

        reverseRune(sRune, i, end)
    }

    ret := string(sRune)

    return ret
}

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