/*Given two strings s and t, write a function to determine if t is an anagram of s. For example, s = "anagram", t = "nagaram", return true. s = "rat", t = "car", return false. Note: You may assume the string contain…

Given two strings s and t, write a function to determine if t is an anagram of s. For example, s = "anagram", t = "nagaram", return true. s = "rat", t = "car", return false. Note: You may assume the string contains…

Array 448.找出数组中所有消失的数 要求:整型数组取值为 1 ≤ a[i] ≤ n,n是数组大小,一些元素重复出现,找出[1,n]中没出现的数,实现时时间复杂度为O(n),并不占额外空间 思路1:(discuss)用数组下标标记未出现的数,如出现4就把a[3]的数变成负数,当查找时判断a的正负就能获取下标 tips:注意数组溢出 public List<Integer> findDisappearedNumbers(int[] nums) { List<Integer> d…

(记忆线:当时一刷完是1-205. 二刷88道.下次更新记得标记不能bug-free的原因.)   88-------------Perfect Squares(完美平方数.给一个整数,求出用平方数来相加得到最小的个数) public class Solution{ public static void main(String[] args){ System.out.println(numSquares(8)); } public static int numSquares(int n){ //…

Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = "anagram", t = "nagaram", return true.s = "rat", t = "car", return false. Note:You may assume the string contains onl…

Problem: Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = "anagram", t = "nagaram", return true.s = "rat", t = "car", return false. Summary: 给出字符串s和t,判断t是不是s的相同字母异序词.…

(1)Valid Anagram 解题思路: 使用一个数组,首先遍历S相应位置加1,然后遍历T,判断此时如果相应位置为零返回FALSE,否则就减一.T遍历完毕后返回true. 代码如下: public class Solution { public boolean isAnagram(String s, String t) { if (s == null || t == null || s.length() != t.length()) { return false; } int[] alpha…

Valid Anagram My Submissions Question Total Accepted: 43694 Total Submissions: 111615 Difficulty: Easy Given two strings s and t, write a function to determine if t is an anagram of s. For example,s = "anagram", t = "nagaram", return t…

class Solution { public: bool isAnagram(string s, string t) { if(t=="") return s==""; map<char,int> m_s; map<char,int> m_t; ; while(i<t.length()){ if(m_t.find(t[i]) != m_t.end()) m_t[t[i]]++; )); i++; } i=; while(i<s.…

class Solution(object):    def isAnagram(self, s, t):        if sorted(list(s.lower()))==sorted(list(t.lower())):            return True        return Falsesol=Solution()print sol.isAnagram(s='anAgram', t='nagaram')…