zoukankan      html  css  js  c++  java
  • leetcode算法之Valid Anagram

    原文算法说明如下:

    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.

    翻译:判断给定的两个字符串是否为打乱了顺序的同一个字符串

    我的java实现:

    import java.util.HashMap;
    public class Solution {
        public boolean isAnagram(String s, String t) {
        	if(s.length() != t.length()){
        		return false;
        	}
        	char[] c1 = s.toCharArray();
        	char[] c2 = t.toCharArray();
        	HashMap<Character,Integer> map1 = new HashMap<Character,Integer>();
        	HashMap<Character,Integer> map2 = new HashMap<Character,Integer>();
        	for(int i=0;i < s.length();i++){
        		if(!map1.containsKey(c1[i])){
        			map1.put(c1[i], 1);
        		}else{
        			map1.put(c1[i], map1.get(c1[i])+1);
        		}
        		if(!map2.containsKey(c2[i])){
        			map2.put(c2[i], 1);
        		}else{
        			map2.put(c2[i], map2.get(c2[i])+1);
        		}
        	}
        	if(map1.entrySet().containsAll(map2.entrySet()) && map2.entrySet().containsAll(map1.entrySet())){
        		return true;
        	}
            return false;
        }
    }
    



  • 相关阅读:
    [UE4]虚幻引擎的C++环境安装
    [UE4]Drop,扔物品
    [UE4]Grab抓取
    [UE4]抓取准备
    [UE4]用Format Text进行调试
    [UE4]Overlap Event 碰撞事件
    [UE4]Skeletal Mesh的碰撞体
    [UE4]模拟物理
    [UE4]Static Mesh的碰撞体
    [UE4]镜像
  • 原文地址:https://www.cnblogs.com/iamconan/p/7383578.html
Copyright © 2011-2022 走看看