难度系数排序，容易题1-10题：
Cosine Similarity new
Fizz Buzz
O(1)检测2的幂次
x的平方根
不同的路径
不同的路径 II
两个字符串是变位词
两个链表的和
中位数
主元素
Cosine Similarity
题目：
Cosine similarity is a measure of similarity between two vectors of an inner product space that measures the cosine of the angle between them. The cosine of 0° is 1, and it is less than 1 for any other angle.
See wiki: Cosine Similarity
Here is the formula:
Given two vectors A and B with the same size, calculate the cosine similarity.
Return 2.0000 if cosine similarity is invalid (for example A = [0] and B = [0]).
Given A = [1, 2, 3], B = [2, 3 ,4].
Return 0.9926.
Given A = [0], B = [0].
Return 2.0000
解题：
Java程序：
class Solution { /** * @param A: An integer array. * @param B: An integer array. * @return: Cosine similarity. */ public double cosineSimilarity(int[] A, int[] B) { // write your code here int ab = 0; double aa = 0.0; double bb = 0.0; if(A.length != B.length) return 2.0000; if(A.length==1 && A[0]==0 || B.length==1 && B[0]==0) return 2.0000; for(int i=0;i<A.length;i++){ ab += A[i] * B[i]; aa += A[i] * A[i]; bb += B[i] * B[i]; } if(aa==0 || bb==0) return 2.0000; aa = Math.sqrt(aa); bb = Math.sqrt(bb); return ab/(aa*bb); } }
Python程序：
class Solution: """ @param A: An integer array. @param B: An integer array. @return: Cosine similarity. """ def cosineSimilarity(self, A, B): # write your code here ab = 0 aa = 0 bb = 0 if(len(A)!=len(B)) : return 2.0000 # if len(A)==1 and A[0] = 0 or len(B)==1 and B[0]=0 : return 2.0000; for i in range(len(A)): ab += A[i]*B[i] aa += A[i]*A[i] bb += B[i]*B[i] if aa==0 or bb==0 : return 2.0000 aa = aa**0.5 bb = bb**0.5 return ab/(aa*bb)
Fizz Buzz
题目：
给你一个整数n. 从 1 到 n 按照下面的规则打印每个数：
- 如果这个数被3整除，打印
fizz. - 如果这个数被5整除，打印
buzz. - 如果这个数能同时被
3和5整除，打印fizz buzz.
比如 n = 15, 返回一个字符串数组：
["1", "2", "fizz", "4", "buzz", "fizz", "7", "8", "fizz", "buzz", "11", "fizz", "13", "14", "fizz buzz"]
解题：
Java程序:
class Solution { /** * param n: As description. * return: A list of strings. */ public ArrayList<String> fizzBuzz(int n) { ArrayList<String> results = new ArrayList<String>(); for (int i = 1; i <= n; i++) { if (i % 15 == 0) { results.add("fizz buzz"); } else if (i % 5 == 0) { results.add("buzz"); } else if (i % 3 == 0) { results.add("fizz"); } else { results.add(String.valueOf(i)); } } return results; } }
Python程序：
class Solution: """ @param n: An integer as description @return: A list of strings. For example, if n = 7, your code should return ["1", "2", "fizz", "4", "buzz", "fizz", "7"] """ def fizzBuzz(self, n): results = [] for i in range(1, n+1): if i % 15 == 0: results.append("fizz buzz") elif i % 5 == 0: results.append("buzz") elif i % 3 == 0: results.append("fizz") else: results.append(str(i)) return results
O(1)检测2的幂次
O(1) Check Power of 2
题目：
用 O(1) 时间检测整数 n 是否是 2 的幂次。
n=4，返回 true;
n=5，返回 false.
O(1) 时间复杂度
解题：
Java程序：
class Solution { /* * @param n: An integer * @return: True or false */ public boolean checkPowerOf2(int n) { // write your code here if(n==1) return true; if(n<=0 || n%2==1) return false; while(n!=0){ n=n/2; if(n==1) return true; if(n%2==1) return false; } return true; } };
原来就是利用到奇数不是2的次幂，进行排除
这里的时间复杂度应该是O(log2n) ，在最坏的情况下n就是2的幂次方，while循环要进行到k次，2^k = n
但是这个也AC了
看到下面程序，利用到若n是2的次幂则，n和n-1的二进制表示对应位置都不一样
class Solution { /* * @param n: An integer * @return: True or false */ public boolean checkPowerOf2(int n) { // write your code here if(n<=0) return false; return (n & (n-1)) == 0 ? true : false; } };
Python程序：
class Solution: """ @param n: An integer @return: True or false """ def checkPowerOf2(self, n): # write your code here if n<=0 : return False if n==1 : return True if n%2==1 : return False while n!=0: n/=2 if n==1 or n==0: return True if n%2==1:return False return True