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1. When we are sorting the numbers we will first find the number of digits in the biggest number.
2. If there are N digits in the biggest number then we will need to perform N number of pass.
3. We will pad the remaining numbers with leading zeros so they all have N digits.
4. Then we will take 10 buckets labeled 0 to 9 and sort the numbers.
Time Complexity : O(N*n) where N is the digits of the biggest number, and n is the length of the array.
Space Complexity: O(1), it will take 10 buckets.
https://www.youtube.com/watch?v=YXFI4osELGU
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原文地址：http://www.cnblogs.com/winscoder/p/4269056.html
