【LeetCode】50. Pow(x, n)

题目：

50. Pow(x, n)

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• Total Submissions: 539052
• Difficulty: Medium
• Contributor: LeetCode

Implement pow(x, n).

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解题思路：pow(x,n)即n个x相乘，用二分的思想，当n为偶数时，可以表示为pow(x,n/2)*pow(x,n/2)，当n为基数时，可以表示为pow(x,n/2)*pow(x,n/2)*x.

解答：

class Solution {
public:
    double myPow(double x, int n) {
        if(n < 0){
        return 1.0/myPow1(x,-n);
}
else{
return myPow1(x,n);
}
    }
    double myPow1(double x,int n){
    if(n == 0){
    return 1.0;
}
double y = myPow1(x,n/2);
if(n & 1){
return y*y*x;
}
else{
return y*y;
}
} 
};