[LeetCode]242. Valid Anagram ★

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本文链接： https://blog.csdn.net/xingyu97/article/details/100159661

题目描述

Given two strings s and t , write a function to determine if t is an anagram of s.
题目大意：给定两个字符串 s 和 t ，判断 t 是否是 s 中的字符变换顺序得到的。如果是返回True，否则返回False

样例

Example 1:

Input: s = “anagram”, t = “nagaram”
Output: true

Example 2:

Input: s = “rat”, t = “car”
Output: false

python解法

class Solution:
    def isAnagram(self, s: str, t: str) -> bool:
        if len(s) != len(t): return False
        d1, d2 = {}, {}
        for i in s:
            if i in d1:d1[i] += 1
            else: d1[i] = 1
        for i in t:
            if  i in d2: d2[i] += 1
            else: d2[i] = 1
        for i in s:
            if i not in t or d1[i] != d2[i]:
                return False
        return True

Runtime: 60 ms, faster than 60.76% of Python3 online submissions for Valid Anagram.
Memory Usage: 13.9 MB, less than 15.63% of Python3 online submissions for Valid Anagram.
题后反思：无

C语言解法

int cmp(const void *a, const void *b)
{
    return *(char*)a - *(char*)b;
}

bool isAnagram(char * s, char * t){
    if (strlen(s)!=strlen(t))
        return false;
    qsort(s, strlen(s), sizeof(char), cmp);
    qsort(t, strlen(s), sizeof(char), cmp);
    for (int i=0;i<strlen(s);i++)
    {
        if (s[i] != t[i])
            return false;
    }
    return true;
}

Runtime: 88 ms, faster than 24.48% of C online submissions for Valid Anagram.
Memory Usage: 7.4 MB, less than 25.00% of C online submissions for Valid Anagram.

题后反思：

1. 测试样例中应该是有较大的的数据，导致使用排序然后依次比较的方法耗时较长。

文中都是我个人的理解，如有错误的地方欢迎下方评论告诉我，我及时更正，大家共同进步