R2-字符串

目录

解简单方程法

线性代数法

ps:

就是从里面找出one,two,zero,---nine

想到哈希表,key代表单词，value代表0---9

用t表示单词，那不就是t在s中的查找问题了吗

但这样显然有些麻烦，在于t是不确定的,t需要遍历一遍keys()，然后再在s中查找，时间复杂度会比较高，应该超时。

解简单方程法

ben神

class Solution:
    def originalDigits(self, s: str) -> str:
        nums=["zero","one","two","three","four","five","six","seven","eight","nine"]
        order=[(0,"z"),(2,"w"),(4,"u"),(5,"f"),(6,"x"),(7,"s"),(1,"o"),(8,"g"),(3,"h"),(9,"i")]
        def reduce(num,v):
            for c in nums[num]:
                cnts[c]-=v
        #返回字典代表每个字符的计数
        cnts=Counter(s)
        ret=[0]*10
        for num,key in order:
            #v代表字符串中特征字符的个数,num是要返回的数字，也作为下标
            v=cnts[key]
            reduce(num,v)
            ret[num]=v
        return "".join(str(i)*v for i,v in enumerate(ret))

线性代数法

. - 力扣（LeetCode）

 

 

class Solution:
    def originalDigits(self, s: str) -> str:
        cnt = Counter(s)
        return "0" * cnt["z"] + \
               "1" * (cnt["o"] - cnt["z"] - cnt["u"] - cnt["w"]) + \
               "2" * cnt["w"] + \
               "3" * (cnt["r"] - cnt["z"] - cnt["u"]) + \
               "4" * cnt["u"] + \
               "5" * (cnt["v"] - cnt["s"] + cnt["x"]) + \
               "6" * cnt["x"] + \
               "7" * (cnt["s"] - cnt["x"]) + \
               "8" * (cnt["t"] - cnt["w"] - cnt["r"] + cnt["z"] + cnt["u"]) + \
               "9" * ((cnt["n"] - cnt["s"] + cnt["x"] - cnt["o"] + cnt["z"] + cnt["u"] + cnt["w"]) // 2)

 

ps:

第2种还没看懂。稍后再说