python動態生成多個變量

最近刷leetcode遇到分行取字符串再重新拼接的題目，解題過程中使用了動態生成變量的相關辦法。后來發現用字典更加簡單。

動態生成如下：

for i in range(5):
    locals()['L%s'%i] = 'haha'
locals()['L%s'%2] = 'xixi'

print locals()['L%s'%1],locals()['L%s'%2],L2


output:
haha xixi xixi

其中運用了locals()來返回變量。

這個方法其實很一般，如果運用迭代來生成字典則非常簡單。

附動態生成變量與字典的截圖代碼對比：

#動態生成變量
class Solution(object):
    def convert(self, s, numRows):
        if numRows == 1:
            return s
        names = locals()
        for i in range(1,numRows+1):
            names['l%s'%i] = ''
        n = 1
        i = 0
        while i < len(s):
            while n < numRows:
                if i == len(s):
                    break
                names['l%s'%n] += s[i]
                n +=1
                i +=1
            while n > 1:
                if i == len(s):
                    break
                names['l%s'%n] += s[i]
                n -= 1
                i += 1
        out = ''
        for i in range(1,numRows+1):
            print names['l%s'%i]
            out +=names['l%s'%i] 
        print out
        return out 


#使用字典
class Solution(object):
    def convert(self, s, numRows):
        if numRows == 1:
            return s
        ldict = {}
        for i in range(1,numRows+1):
            ldict[i] = ''
        n = 1
        i = 0
        while i < len(s):
            while n < numRows:
                if i == len(s):
                    break
                ldict[n] += s[i]
                n +=1
                i +=1
            while n > 1:
                if i == len(s):
                    break
                ldict[n] += s[i]
                n -= 1
                i += 1
        out = ''
        for i in ldict.values():
            out +=i 
        return out 

字典顯得簡潔的多，更適合python。