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阿新 • • 發佈:2020-08-16
實現函式double Power(double base, int exponent),求base的exponent次方。不得使用庫函式,同時不需要考慮大數問題。
示例 1:
輸入: 2.00000, 10
輸出: 1024.00000
示例2:
輸入: 2.10000, 3
輸出: 9.26100
示例3:
輸入: 2.00000, -2
輸出: 0.25000
解釋: 2-2 = 1/22 = 1/4 = 0.25
說明:
-100.0 <x< 100.0
n是 32 位有符號整數,其數值範圍是[−231,231− 1] 。
code1:迭代
class Solution { public: double myPow(doublex, int n) { if(n == 1 || n == 0) { return n == 1 ? x : 1; } int sign = 1; long num = n; if (num < 0) { sign = -1; num = -num; } double res = 1; while (num) { if (num & 1) { res*= x; } x *= x; num >>= 1; } return sign == 1 ? res : 1 / res; } };
code2:遞迴
class Solution { private: double myPowCore(double x, int n) { if (n == 0) { return 1; } double res = myPowCore(x, n/2);if(n & 1) { return res * res * x; } else { return res * res; } } public: double myPow(double x, int n) { if(n == 1 || n == 0) { return n == 1 ? x : 1; } if(n > 0){ return myPowCore(x, n); } return myPowCore(1/x, n); } };