注意边界情况：当n==INT_MIN时，-INT_MIN超出范围了，要改为INT_MAX，
又因为存在（-1，-INT_MIN）这个特殊情况存在，考虑到正负修正为INT_MAX-1
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| class Solution {
public:
double myPow(double x, int n) {
if(n < 0)
return myPow(1 / x, n == INT_MIN ? INT_MAX - 1: -n);
return fast(x, n);
}
double fast(double x, int n) {
if(n == 0)
return 1;
double half = fast(x, n /2);
if(n % 2) {
return half*half*x;
}
return half*half;
}
};
|
Accepted
304/304 cases passed (4 ms)
Your runtime beats 71.28 % of cpp submissions
Your memory usage beats 57.93 % of cpp submissions (8.3 MB)
