统计阶乘之后的结果有多少个0,就是统计阶乘有多少个5 n! = N*(N-1)*(N-2)*(N-3)...... 1...5....2*5 .....4*5 .....5*5.....5*5.......2*5*5......5*5*5......n count = n/5+n/25+n/125+..... public int trailingZeroes(int n) { int count = 0; for (int i = 1; i <= n; i++) { int N = i...