DAY 8

344. 反转字符串

朴实无华

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func reverseString(s []byte) {
    n := len(s)

    for i := 0; i < n/2; i++ {
        s[n-i-1], s[i] = s[i], s[n-i-1]
    }
}

541.反转字符串 II

朴实无华的递归

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func reverseStr(s string, k int) string {
    n := len(s)
    if n <= k {
        return reverse(s)
    } else if n > k && n < 2*k {
        return reverse(s[:k]) + s[k:]
    } else {
        return reverse(s[:k]) + s[k:2*k] + reverseStr(s[2*k:], k)
    }

}

func reverse(s string) string {
    bytes := []byte(s)
    n := len(bytes)
    for i := 0; i < n/2; i++ {
        bytes[i], bytes[n-i-1] = bytes[n-i-1], bytes[i]
    }
    return string(bytes)
}

LCR 122.路径加密

朴实无华

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func pathEncryption(path string) string {
	ans := ""
	for _, c := range path {
		if c == '.' {
			ans += " "
		} else {
			ans += string(c)
		}
	}
	return ans
}

151.反转字符串中的单词

Go 写起来真舒服

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func reverseWords(s string) string {
    for s[0] == ' ' {
        s = s[1:]
    }
    n := len(s)
    ans := ""

    for i := 0; i < n; i++ {
        word := ""
        for i < n && s[i] != ' ' {
            word += string(s[i])
            i++
        }
        ans = " " + word + ans

        for i+1 < n && s[i+1] == ' ' {
            i++
        }
    }
    return ans[1:]
}

LCR 182.动态口令

？？？

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func dynamicPassword(password string, target int) string {
    return password[target:] + password[:target]
}

DAY 9

28.找出字符串中第一个匹配项的下标

简单复习了一下 KMP，但你要我写我肯定写不出来

这东西写法太多而且感觉有点乱

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func strStr(haystack string, needle string) int {
    n, m := len(haystack), len(needle)
    i, j := 0, 0
    next := getNext(needle)
    for i < n {
        if j == -1 || haystack[i] == needle[j] {
            i++
            j++
        } else {
            j = next[j]
        }
        if j == m {
            j = next[j]
            return i - m
        }
    }
    return -1
}

func getNext(s string) []int {
    n := len(s)
    next := make([]int, n+1)
    i, j := 0, -1
    next[0] = -1
    for i < n {
        if j == -1 || s[i] == s[j] {
            i++
            j++
            next[i] = j
        } else {
            j = next[j]
        }
    }
    return next
}

459. 重复的子字符串

真的是非常巧妙的算法

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func repeatedSubstringPattern(s string) bool {
    return strings.Contains(s[1:]+s[:len(s)-1], s)
}