python學習（十九）

注意，答案只是代表是他人寫的代碼，正確，但不一定能通過測試（比如超時），列舉出來只是它們擁有着獨到之處，雖然大部分確實比我的好
（發現最近光寫解答沒寫思路了）

344. Reverse String

題目

Write a function that takes a string as input and returns the string reversed.

Example:
Given s = “hello”, return “olleh”.

思路與解答

這道題我以前沒做過嘛？

return s[::-1]

答案

這個方案是分治

class Solution(object):
    def reverseString(self, s):
        l = len(s)
        if l < 2:
            return s
        return self.reverseString(s[l/2:]) + self.reverseString(s[:l/2])

原地互換法，可是還是有新列表了啊

class SolutionClassic(object):
    def reverseString(self, s):
        r = list(s)
        i, j  = 0, len(r) - 1
        while i < j:
            r[i], r[j] = r[j], r[i]
            i += 1
            j -= 1

        return "".join(r)

emmm

return ''.join(reversed(s))

541. Reverse String II

題目

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

思路與解答

描述上覆雜了一些
經過測試我發現我理解錯題目了，擦
根據條件不同採取不同的方法

        l,i = len(s),len(s)
        r,nr= list(s),[]
        while i > 0:
            if i >= 2*k:
                nr += list(reversed(r[l-i:l-i+k])) + r[l-i+k:l-i+2*k]
                i = i - 2*k   
                if i <= 0:return "".join(nr)
            elif k <= i < 2*k:
                nr += list(reversed(r[l-i:l-i+k])) + r[l-i+k:]
                return "".join(nr)
            elif i < k:
                nr += list(reversed(r[l-i:]))
                return "".join(nr)

感覺有非常大的優化空間

        l,i = len(s),len(s)
        r,nr= list(s),[]
        while i > 0:
            if i >= 2*k:
                nr += list(reversed(r[l-i:l-i+k])) + r[l-i+k:l-i+2*k] 
            elif k <= i < 2*k:
                nr += list(reversed(r[l-i:l-i+k])) + r[l-i+k:]
            elif i < k:
                nr += list(reversed(r[l-i:]))
            i = i - 2*k
        return "".join(nr) 

答案

牛逼，看穿了題目要求

def reverseStr(self, s, k):
    s = list(s)
    for i in xrange(0, len(s), 2*k):
        s[i:i+k] = reversed(s[i:i+k])
    return "".join(s)

最近好多遞歸，厲害

       return s[:k][::-1] + s[k:2*k] + self.reverseStr(s[2*k:], k) if s else ""

同樣是一行，讀起來有些費勁

return "".join(s[i*k: i*k+k][::(i&1)*2-1] for i in xrange(len(s)/k + 1))

345. Reverse Vowels of a String

題目

Write a function that takes a string as input and reverse only the vowels of a string.

Example 1:
Given s = “hello”, return “holle”.

Example 2:
Given s = “leetcode”, return “leotcede”.

Note:
The vowels does not include the letter “y”.

思路與解答

題目是什麼意思？？？
試了一下是隻反轉字符串中的aeiou？
指定字符的反轉，怎麼寫呢
找清位置，逐個反轉

        r = "aeiouAEIOU"
        d = {}
        ns =list(s)
        for x in r:
            d[x] = x
        l = [i for i,j in enumerate(s) if j in d]
        i,j = 0,len(l)-1
        while i < j:
            ns[l[i]],ns[l[j]]=ns[l[j]],ns[l[i]]
            i += 1
            j -= 1
        return "".join(ns)

答案

這寫的啥呀

    return re.sub('(?i)[aeiou]', lambda m, v=re.findall('(?i)[aeiou]', s): v.pop(), s)

感覺不如我的，思想上差不多

class Solution(object):
    def reverseVowels(self, s):
        vowels = set(list("aeiouAEIOU"))
        s = list(s)
        ptr_1, ptr_2 = 0, len(s) - 1
        while ptr_1 < ptr_2:
            if s[ptr_1] in vowels and s[ptr_2] in vowels:
                s[ptr_1], s[ptr_2] = s[ptr_2], s[ptr_1]
                ptr_1 += 1
                ptr_2 -= 1
            if s[ptr_1] not in vowels:
                ptr_1 += 1
            if s[ptr_2] not in vowels:
                ptr_2 -= 1
        return ''.join(s)

13. Roman to Integer

題目

Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

思路與解答

這題。。。羅馬數字我就會從一數到五
規則蠻複雜的

希望正規點，沒有第三條
提交過程中發現了第三條，啊啊啊啊啊

        d = {'M':1000, 'D':500, 'C':100, 'L':50, 'X':10, 'V':5, 'I':1}
        sums = 0
        ls =list(s)
        for i in range(len(ls)-1):
            if d[ls[i]] < d[ls[i+1]]:
                sums -= d[ls[i]]
            else:
                sums += d[ls[i]]
        return sums + d[ls[-1]]

第三條規則只要判斷一下左邊的是否比右邊的小即可

答案

差不多

d = {"I":1, "V":5, "X":10, "L":50, "C":100, "D":500, "M":1000}
        res = 0
        pre = 0
        for ch in s[::-1]:
            num = d[ch]
            if num >= pre:
                res += num
            else:
                res -= num
            pre = num
        return res

189. Rotate Array

題目

Rotate an array of n elements to the right by k steps.

For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4].

Note:
Try to come up as many solutions as you can, there are at least 3 different ways to solve this problem.

[show hint]

Hint:
Could you do it in-place with O(1) extra space?
Related problem: Reverse Words in a String II

思路與解答

沒太懂題的意思
就是把最後k個字符移到前面來？

        while k > 0:
            nums.insert(0,nums.pop())
            k -= 1

5%啊。。。
直接用切片？主要是我忘了怎麼用切片修改原list了

答案

哦對，是nums[:]，我用nums當然修改不了

        k = k % len(nums)
        nums[:] = nums[-k:] + nums[:-k]

話說是不是我忘記給k取餘所以太慢了？
。。。。並不是

雖然不是python ，但是很有趣。。。n是啥，nums長度？，這個函數又是啥，爲什麼要兩個參數，爲什麼nums還加上n？和我預期的好像不太一樣，算了
第一行反轉所有元素，第二個反向k％n個元素需要旋轉，第三個反向剩餘。如果k大於n，則與k％n相同。你最好以一個例子來想

void rotate(int nums[], int n, int k) {
    reverse(nums,nums+n);
    reverse(nums,nums+k%n);
    reverse(nums+k%n,nums+n);
}