力扣题-11.29

[力扣刷题攻略] Re：从零开始的力扣刷题生活

力扣题1：032. 有效的字母异位词

解题思想：直接遍历即可

class Solution(object):def isAnagram(self, s, t):""":type s: str:type t: str:rtype: bool"""if s==t:return Falseif len(s)!=len(t):return Falsechar_count = {}for i in range(len(s)):if s[i] in char_count:char_count[s[i]] += 1else:char_count[s[i]] = 1for i in range(len(t)):if t[i] in char_count:char_count[t[i]] -=1if char_count[t[i]]<0:return Falseelse:return Falsereturn True

class Solution {
public:bool isAnagram(string s, string t) {if (s == t) {return false;}if (s.length() != t.length()) {return false;}std::unordered_map<char, int> char_count;for (int i = 0; i < s.length(); ++i) {if (char_count.find(s[i]) != char_count.end()) {char_count[s[i]] += 1;} else {char_count[s[i]] = 1;}}for (int i = 0; i < t.length(); ++i) {if (char_count.find(t[i]) != char_count.end()) {char_count[t[i]] -= 1;if (char_count[t[i]] < 0) {return false;}} else {return false;}}return true;}
};