[精选代码笔记]Anagram, group-anagrams, two sum

Valid anagram

242. 有效的字母异位词
题目链接:https://leetcode-cn.com/problems/valid-anagram/
给定两个字符串 s 和 t ，编写一个函数来判断 t 是否是 s 的字母异位词。示例 1:输入: s = "anagram", t = "nagaram"
输出: true
示例 2:输入: s = "rat", t = "car"
输出: false
说明:
你可以假设字符串只包含小写字母。进阶:
如果输入字符串包含 unicode 字符怎么办？你能否调整你的解法来应对这种情况？

Python

# 思路：手动模拟hashtable，将字符串”a-z“的ASCII码作key，计数求差异    def  isAnagram(self, s: str, t: str) -> bool:arr1, arr2 = [0]*26, [0]*26for i in s:arr1[ord(i) - ord('a')] += 1for i in t:arr2[ord(i) - ord('a')] += 1return arr1 == arr2

# 思路：map计数，对比计数差异def isAnagram(self, s: str, t: str) -> bool:dict1, dict2 = {}, {}for item in s:dict1[item] = dict1.get(item,0) + 1for item in t:dict2[item] = dict2.get(item,0) + 1return dict1 == dict2

# 思路：数组排序后比较差异def isAnagram(self, s: str, t: str) -> bool:return sorted(s) == sorted(t)

Java

public class Solution {public boolean isAnagram(String s, String t) {if(s.length() != t.length()) return false;int [] a = new int [26];for(Character c : s.toCharArray()) a[c - 'a']++;for(Character c : t.toCharArray()) {if(a[c -'a'] == 0) return false;a[c - 'a']--;}return true;}
}

public boolean isAnagram(String s1, String s2) {int[] freq = new int[256];for(int i = 0; i < s1.length(); i++) freq[s1.charAt(i)]++;for(int i = 0; i < s2.length(); i++) if(--freq[s2.charAt(i)] < 0) return false;return s1.length() == s2.length();
}

public boolean isAnagram(String s, String t)
{char[] sChar = s.toCharArray();char[] tChar = t.toCharArray();Arrays.sort(sChar);Arrays.sort(tChar);return Arrays.equals(sChar, tChar);
}

Group Anagrams:

49. 字母异位词分组
题目链接：https://leetcode-cn.com/problems/group-anagrams/
给定一个字符串数组，将字母异位词组合在一起。字母异位词指字母相同，但排列不同的字符串。示例:输入: ["eat", "tea", "tan", "ate", "nat", "bat"]
输出:
[["ate","eat","tea"],["nat","tan"],["bat"]
]
说明：所有输入均为小写字母。
不考虑答案输出的顺序。

Python

def groupAnagrams(self, strs):d = {}for w in sorted(strs):key = tuple(sorted(w))d[key] = d.get(key, []) + [w]return d.values()

def groupAnagrams(self, strs):dic = {}for item in sorted(strs):sortedItem = ''.join(sorted(item))dic[sortedItem] = dic.get(sortedItem, []) + [item]return dic.values()

Java

public List<List<String>> groupAnagrams(String[] strs) {List<List<String>> res = new ArrayList<>();HashMap<String, List<String>> map = new HashMap<>();Arrays.sort(strs);for (int i = 0; i < strs.length; i++) {String temp = strs[i];char[] ch = temp.toCharArray();Arrays.sort(ch);if (map.containsKey(String.valueOf(ch))) {map.get(String.valueOf(ch)).add(strs[i]);} else {List<String> each = new ArrayList<>();each.add(strs[i]);map.put(String.valueOf(ch), each);}}for (List<String> item: map.values()) {res.add(item);}return res;
}

Two sum

1. 两数之和
题目链接：https://leetcode-cn.com/problems/two-sum/description/
给定一个整数数组 nums 和一个目标值 target，请你在该数组中找出和为目标值的那 两个 整数，并返回他们的数组下标。
你可以假设每种输入只会对应一个答案。但是，数组中同一个元素不能使用两遍。示例:给定 nums = [2, 7, 11, 15], target = 9因为 nums[0] + nums[1] = 2 + 7 = 9
所以返回 [0, 1]

Python

    def twoSum(self, nums, target):d = dict()for index,num in enumerate(nums):if d.get(num) == None:d[target - num] = indexelse:return [d.get(num), index]

Java

public int[] twoSum(int[] nums, int target) {HashMap<Integer, Integer> tracker = new HashMap<Integer, Integer>();int len = nums.length;for (int i = 0; i < len; i++){if (tracker.containsKey(nums[i])){int left = tracker.get(nums[i]);return new int[]{left+1, i+1};} else {tracker.put(target - nums[i], i);}}return new int[2];}

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[精选代码笔记]Anagram, group-anagrams, two sum

Valid anagram

Python

Java

Group Anagrams:

Python

Java

Two sum

Python

Java

[精选代码笔记]Anagram, group-anagrams, two sum相关推荐

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