[leetcode] Pow(x, n)

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From : https://leetcode.com/problems/powx-n/

Implement pow(x, n).

递归求解，空间复杂度为O(logn)，时间复杂度为O(logn)

class Solution {public:    double myPow(double x, int n) {        long long int num = n;        if(x==0.0) return 0;        if(x == 1) return 1;        if(x == -1) return (!(num&1))-(num&1);        if(num < 0) return myPowCore(1/x, -num);        return myPowCore(x, num);    }        double myPowCore(double x, long long int n) {        if(n == 0) return 1;        if(n == 1) return x;        return ((n&1)*x+(!(n&1)))*myPowCore(x*x, n>>1);            }};

非递归方法，空间复杂度为O(1)，时间复杂度为O(logn)

class Solution {public:    double myPow(double x, int n) {        long long m = n;        if(x == 0) return 0;else if(m == 0 || x == 1) return 1;else if(m < 0) {m = -m;x = 1/x;}double ans = 1;while(m) {if(m&1) {ans *= x;}m = m>>1;x *= x;}return ans;    }};

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