Given an integer n, return the number of trailing zeroes in n!.
Note: Your solution should be in logarithmic time complexity.
⌊ n/2 ⌋
times.[3, 30, 34, 5, 9]
, the largest formed number is 9534330
.A -> 1 B -> 2 C -> 3 ... Z -> 26 AA -> 27 AB -> 28
1 -> A 2 -> B 3 -> C ... 26 -> Z 27 -> AA 28 -> AB
gas[i]
.cost[i]
of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.1,2,3
→ 1,3,2
3,2,1
→ 1,2,3
1,1,5
→ 1,5,1
"PAYPALISHIRING"
is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)P A H N A P L S I I G Y I RAnd then read line by line:
"PAHNAPLSIIGYIR"
string convert(string text, int nRows);
convert("PAYPALISHIRING", 3)
should return "PAHNAPLSIIGYIR"
.
gas[i]
.cost[i]
of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.[1,2,0]
return 3
,[3,4,-1,1]
return 2
.null
.