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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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