11. Sorting
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2 for loops
every time compare and swap 2 adjacent elements
find the largest element and move to the end
Time: O(N^2) and space: O(1)
2 for loops
every loop find the smallest element and move to the beginning
split the arrays into 2 parts, sorted subarray and unsorted subarray
Time: O(N^2) and space: O(1)
给一个int[] nums, 返回一个用nums里的num组成的最大的数
custom sort: basically implement a String comparator to decide which String should come first during concatenation