557.反转字符串中的单词

一、暴力

看到第一想法，首先就是找到一个空格，将空前的那个单词反转。利用了一个for循环，对于每个单词还有一个while循环。速度感人。

执行用时: 440 ms	在Reverse Words in a String III的Python3提交中击败了0.84% 的用户
内存消耗: 14.2 MB	在Reverse Words in a String III的Python3提交中击败了0.62% 的用户

class Solution:
    def reverseWords(self, s):
        x = 0
        num = []
        for i in s:
            num.append(i)
        for i in range(len(s)):
            if  i == len(s)-1:
                while x < i:
                    num[x],num[i] = num[i],num[x]
                    x += 1
                    i -= 1
            if  num[i] == ' ' :
                j = i
                r = i
                while x < j-1:
                    num[x],num[j-1] = num[j-1],num[x]
                    x += 1
                    j -= 1
                x = r+1
        s = ''
        for i in num:
            s += i        
        return s 

二、暴力法改进

改成while循环莫名变快了，而且每次提交速度都不一样，真神奇。。。

执行用时: 244 ms	在Reverse Words in a String III的Python3提交中击败了1.35% 的用户
内存消耗: 13.7 MB,	在Reverse Words in a String III的Python3提交中击败了0.62% 的用户

class Solution:
    def reverseWords(self, s):
        x = 0
        num = []
        for i in s:
            num.append(i)
        i = 0
        while i < len(s):
            if  i == len(s)-1:
                while x < i:
                    num[x],num[i] = num[i],num[x]
                    x += 1
                    i -= 1
                break
            if  num[i] == ' ':
                j = i
                r = i
                while x < j-1:
                    num[x],num[j-1] = num[j-1],num[x]
                    x += 1
                    j -= 1
                x = r+1
                i = x+1
            else:
                i += 1
               
        s = ''
        for i in num:
            s += i        
        return s       

三、切片

执行用时: 200 ms	在Reverse Words in a String III的Python3提交中击败了2.03% 的用户
内存消耗: 14.1 MB,	在Reverse Words in a String III的Python3提交中击败了0.62% 的用户

class Solution:
    def reverseWords(self, s):
        num = []
        k = 0
        for i in s:
            num.append(i)
        i = 0
        while i < len(num):
            if  i == len(s)-1:
                if k == 0:
                    num[::]=num[::-1]
                else:
                    num[k::] = num[i:k-1:-1]
                break
            if  num[i] == ' ':
                if k == 0:
                    num[:i:]=num[i-1::-1]
                else:
                    num[k:i:]=num[i-1:k-1:-1]
                k = i+1
            i += 1   
        s = ''
        for i in num:
            s += i        
        return s