[LeetCode]Factorial Trailing Zeroes
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Given an integer n, return the number of trailing zeroes inn!.
Note: Your solution should be in logarithmic time complexity.
求n的阶乘后面有几个零,要求时间复杂度O(logn)。
刚看到这个题有点蒙,后来想了想,零是怎么增加的,只有10*10才会增加,阶乘中从不缺偶数,所以n中有几个5的因子就有几个零,5!只有1个零,10!有2个零,11-14也只有2个零,15!就有三个零了。
int trailingZeroes(int n) { int num = 0; while(n) { n /= 5; num += n; } return num; }时间复杂度O(logn) ,响应时间0ms。
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