Pow(x, n)
Total Accepted: 73922 Total Submissions: 269855 Difficulty: Medium
Implement pow(x, n).
 1.递归
 
/* n = 0, <0 , >0 x = 0 ,x>0,x<0 */ class Solution { public: double myPowWithPositiveExp(double x,unsigned int n) { if(n==1){ return x; } double res = myPowWithPositiveExp(x,n>>1) ; res = res * res; return 1&n ? res * x : res; } double myPow(double x, int n) { if(n==0){ return 1; } if(n<0){ return 1/ myPowWithPositiveExp(x,-n) ; } return myPowWithPositiveExp(x,n) ; } };
2.迭代
/* n = 0, <0 , >0 x = 0 ,x>0,x<0 */ class Solution { public: double myPow(double x, int n) { if(n==0){ return 1; } long long int m = fabs(n); double res = 1; bool first = true; while(m){ first ? x *= 1,first=false : x *= x; if(1&m){ res *= x; } m = m>>1; } return n<0 ? 1/res : res; } };
Next challenges: (M) Sqrt(x)
写者:zengzy
出处: http://www.cnblogs.com/zengzy
标题有【转】字样的文章从别的地方转过来的,否则为个人学习笔记
 
                     
                    
                 
                    
                 
                
            
         
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浙公网安备 33010602011771号