[LeetCode#179]Largest Number

The problem is also a new kind of problem!!!
The solution is a magic!!!
Wrong direction: try to use the digit to arrange those numbers. This method is very problemetic, since not all nums in the same length!!! Compare which digit???

A more smart way is to take advantage the sorting in the language, rather than to design our own version.
Key: since those nums were designed to concatenate with each other, rather than add. 
Besides the way to multiply a number: num1 * 100 + num2
Why not we directly use String concatenation: str1 + str2.

To make things better, since all the nums only include digits, we can directly compare the string representation of two numbers:
nums1 < nums2  <====> nums1.toString < num2.toString

How to arrange the concatenate order to make the result string be the largest?
Define a comparator to sort strings, make the decision based on the concatnated string rather than s1 or s2. 
Arrays.sort(strs, new Comparator<String> () {
    public int compare(String s1, String s2) {
        String temp1 = s1 + s2;
        String temp2 = s2 + s1;
        return -1 * temp1.compareTo(temp2);
    }
});

After the sort, we rank the first nums as the num would bigger than any other concatnnation. 

Also, the result the string may start with 0.
Case: [0, 0, 0, 0]
We should only keep one 0 if the answer is 0.

while (buffer.length() > 1 && buffer.charAt(0) == ‘0‘)
    buffer.deleteCharAt(0);
return buffer.toString();


Key:
This problem is not logic hard, but requires some different ideas in comparison and sorting.

public class Solution {
    public String largestNumber(int[] nums) {
        if (nums == null || nums.length == 0)
            return "";
        String[] strs = new String[nums.length];
        for (int i = 0; i < nums.length; i++)
            strs[i] = String.valueOf(nums[i]);
        Arrays.sort(strs, new Comparator<String> () {
            public int compare(String s1, String s2) {
                String temp1 = s1 + s2;
                String temp2 = s2 + s1;
                return -1 * temp1.compareTo(temp2);
            }
        });
        StringBuffer buffer = new StringBuffer();
        for (int i = 0; i < strs.length; i++) {
            buffer.append(strs[i]);
        }
        while (buffer.length() > 1 && buffer.charAt(0) == ‘0‘)
            buffer.deleteCharAt(0);
        return buffer.toString();
    }
}