LeetCode算法题（C#版）

两数之和

方法一：暴力枚举

class Solution {public int[] TwoSum(int[] nums, int target) {for(int i=0;i<nums.Length;i++){for(int j=i+1;j<nums.Length;j++){if(nums[i]+nums[j]==target){return new int[]{i,j};}}}return new int[0];}
}

方法二：C#知识中的字典

class Solution {public int[] TwoSum(int[] nums, int target) {Dictionary<int, int> hashtable = new Dictionary<int, int>();for (int i = 0; i < nums.Length; ++i) {if (hashtable.ContainsKey(target - nums[i])) {return new int[]{hashtable[target - nums[i]], i};}hashtable[nums[i]]=i;}return new int[0];}
}

回文数

代码

public class Solution {public bool IsPalindrome(int x) {int revertedNumber=0;if(x<0||(x>0&&x%10==0)){return false;}while(x>revertedNumber){revertedNumber=revertedNumber*10+x%10;x=x/10;}return x==revertedNumber||x==revertedNumber/10;}
}

罗马数组转整数

代码

public class Solution {public int RomanToInt(string s) {Dictionary<char,int> romanToInt=new Dictionary<char,int>();romanToInt.Add('I',1);romanToInt.Add('V',5);romanToInt.Add('X',10);romanToInt.Add('L',50);romanToInt.Add('C',100);romanToInt.Add('D',500);romanToInt.Add('M',1000);int answer=0;for(int i=0;i<s.Length;i++){int value=romanToInt[s[i]];if(i<s.Length-1&&value<romanToInt[s[i+1]]){answer-=value;}else{answer+=value;}}return answer;}
}

构成整天的下标对数目I

代码

和两数之和的逻辑相同

public class Solution {public int CountCompleteDayPairs(int[] hours) {int sum=0;for(int i=0;i<hours.Length;i++){for(int j=i+1;j<hours.Length;j++){if((hours[i]+hours[j])%24==0){sum++;}}}return sum;}
}