publicclassSolution{//递归publicdoubleMyPow(double x,int n){return(n>=0)?res(x,n):1/res(x,n);}privatedoubleres(double x,int n){if(n==0)return1;double y=res(x,n/2);return(n%2==0)?y*y:y*y*x;}} 3、时间复杂度 时间复杂度 :O(
leetcode--Power(x,n) Implement pow(x,n). divide and conquer method: the running time is O(ln n) 1publicclassSolution{2publicdoublepow(doublex,intn){3doubleresult = 0;4longtemp =n;5if(temp >= 0)6result =power(x, temp);7else8result = 1.0 / power(x, -1 *temp);910returnresult...
1 class Solution { 2 3 public double myPow(double x, int n) { 4 // 迭代算法,利用二进制位 5 if(x == 0){ // 0 的任何次方都等于 0,1 的任何次方都等于 1 6 return x; 7 } 8 9 long power = n; // 为了保证-n不溢出,先转换成long类型 10 if(n < 0){ // 如果n小于0, 求1...
Implement pow(x, n), which calculates x raised to the power n (xn). Example 1: Input: 2.00000, 10 Output: 1024.00000 Example 2: Input: 2.10000, 3 Output: 9.26100 Example 3: Input: 2.00000, -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25 Note: -100.0 < x < 100.0 n...
实现pow(x, n) ,即计算 x 的整数 n 次幂函数(即,xn )。 示例1: 输入:x = 2.00000, n = 10 输出:1024.00000 示例2: 输入:x = 2.10000, n = 3 输出:9.26100 示例3: 输入:x = 2.00000, n = -2 输出:0.25000 解释:2-2 = 1/22 = 1/4 = 0.25 提示: -100.0 < x < 100.0 -231 <...
LeetCode50. Pow(x, n) Implementpow(x, n), which calculatesxraised to the powern(i.e.,xn). Example 1: Input: x = 2.00000, n = 10 Output: 1024.00000 Example 2: Input: x = 2.10000, n = 3 Output: 9.26100 Example 3: Input: x = 2.00000, n = -2...
class Solution { public: int mySqrt(int x) { double g=1; while(1){ double g2=(g+x/g)/2; if(abs(g2-g)<0.1)return (int)g2; else g=g2; } return g; } }; 7、如何实现Pow(x,n):Pow(x, n) - LeetCode Implement pow(x, n), which calculates x raised to the power n...
def isPowerOfThree(self, n: int) -> bool: while n and not n % 3: n //= 3 return n == 1 return n > 0 and not 3**19 % n 1. 2. 3. 4. 5. 6. 7. 342. 4的幂 Leetcode 方法一:二进制表示中 1 的位置 如果n 是 4 的幂,那么 n 的二进制表示中有且仅有一个 1,并且这...
x | x == x; 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. //n & (n-1)可以去除n的位级表示中最低的那一位 n = 11110100 n & (n-1) == 11110000 //n & (-n)可以得到n的位级表示中最低的那一位 n = 11110100 n & (-n) == 00000100 ...
0231 Power of Two Go 43.7% Easy 0232 Implement Queue using Stacks Go 49.5% Easy 0233 Number of Digit One 31.3% Hard 0234 Palindrome Linked List Go 39.2% Easy 0235 Lowest Common Ancestor of a Binary Search Tree Go 49.9% Easy 0236 Lowest Common Ancestor of a Binary Tree Go 45.6%...