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(log n) 其中n为遍历的层数。 空间复杂度:O(l...
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...
Implement pow(x, n). 1 class Solution { 2 public: 3 double power(double x, int n) 4 { 5 if (n == 0) 6 return 1; 7 ...
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...
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...
LeetCode50 Power(x,n) leetcode-cn.com/problem 「快速幂算法」的本质是分治算法当我们要计算 x^n时,我们可以先递归地计算出 y = x^[n/2],其中[a]表示对 a 进行下取整;根据递归计算的结果,如果 n为偶数,那么 x^n = y^2如果n 为奇数,那么 x^n =y^2×x;递归的边界为 n = 0,任意数的 0 ...
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 ...
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import java.util.ArrayList;import java.util.LinkedList;import java.util.List;/** * Given a set of distinct integers, nums, return all possible subsets (the * power set). * * Note: The solution set must not contain duplicate subsets. * * Example: * * Input: nums = [1,2,3] Output...
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