[Leetcode]75. Pow (x,n) power operation

Implement Pow (x, n).

Subscribe to see which companies asked this question

Solution 1: The simplest is that n x is multiplied directly, without a doubt time Limit exceeded

classSolution { Public:    DoubleMypow (DoubleXintN) {if(X <0.000001)return 0; if(n = =0)return 1; if(N <0)return 1.0/MYPOW (x,-N); Doubleres =1.0;  for(inti =0; I < n; ++i) Res*=x; returnRes; }};

Solution 2: Consider POW (4,12), can be calculated as: Pow (4,12) =pow (4,8) *pow (bis), and POW (4,8) can be through POW (4,1)-->pow (4,2)-->pow ()-->pow ( 4,8) The same can be calculated by the POW (4,1)-->pow (4,2)-->pow (bis). That is, when calculating to a power pow (x,k), if the current power of the Square pow (x,2k) will not be greater than the result (judging by 2k<=n), then the direct square. So you can write the following code:

classSolution { Public:    DoubleMypow (DoubleXintN) {if(X <0.000001)return 0; if(n = =0)return 1; if(N <0)return 1.0/MYPOW (x,-N); Doubleres =1.0;  while(N >0) {            Doublef =x; intloop =1;  while(Loop <<1) <=N) {f*=F; Loop<<=1; } N-=Loop; Res*=F; }        returnRes; }};

This reduces the number of operations to some extent, but as the above example shows, there are still a lot of redundant computations. Also time Limit exceeded is timed out after Leetcode is committed . You can use map<double,int> to record a calculated power value, and all values after the first loop can be obtained by looking up the table, which reduces the number of duplicate calculations.

classSolution { Public:    DoubleMypow (DoubleXintN) {if(x = =0)return 0; if(n = =0)return 1; if(N <0)return 1.0/MYPOW (x,-N); Vector<Double>Power; Doubleres =1.0; Doublef =x; intloop =1;        Power.push_back (f);  while(Loop <<1) <=N) {f*=F;            Power.push_back (f); Loop<<=1; } N-=Loop; Res*=F; Bitset<sizeof(int) *8>POWB (n);  for(inti =0; I < +; ++i) {if(Powb[i] = =1) Res*=Power[i]; }        returnRes; }};

but this approach will appear Memory Limit exceeded. Because the input exponent is of type int, the actual vector<double> maximum length is 32, which is the constant complexity. The complexity of time is also at the constant level. I do not know what level of space complexity the topic limits.

Solution 3: Consider the use of binary search. If n is an even number, then POW (x,n) =pow (X,N/2) *pow (X,N/2), and if n is an odd number, multiply by x. So the recursion goes down until the n==0.

classSolution { Public:    DoubleMypow (DoubleXintN) {if(x = =0)return 0; if(n = =0)return 1; if(N <0)return 1.0/Power (x,-N); returnPower (x, N); }Private:    DoublePowerDoubleXintN) {if(n = =0)return 1; DoubleHalf = Power (x, N/2); Doubleres = half *half; returnN%2==0? Res:res *x; }};

[Leetcode]75. Pow (x,n) power operation