Java – no suitable method for sort (int [], >) found

Algorithm problem: given a list of nonnegative integers, arrange them into the largest number

For example, given [3,30,34,5,9], the maximum number of formations is 9534330

Note: the result can be very large, so you need to return a string instead of an integer

public class Solution {
    public String largestNumber(int[] num) {
        Arrays.sort(num,new java.util.Comparator<Integer>() {
            @Override
            public int compare(Integer a,Integer b) {
                String s1 = String.valueOf(a),s2 = String.valueOf(b);
                return Integer.parseInt(s1 + s2) - Integer.parseInt(s2 + s1);
            }
        });
        StringBuilder builder = new StringBuilder();
        for (int i = num.length - 1; i >= 0; i--)   builder.append(num[i]);
        return builder.toString();
    }
}

Result: Line 3: error: no suitable method was found for sort (int [], < anonymous comparator < integer > >)

Who knows how to modify it? thank you!

Thank you for all your detailed answers. I have modified my code to

public String largestNumber(int[] num) {
    int N = num.length;
    String[] aux = new String[N];
    for (int i = 0; i < N; i++) aux[i] = String.valueOf(num[i]); // int[] -> String[]
    Arrays.sort(aux,new Comparator<String>() {
        @Override
        public int compare(String a,String b) {
            return (int) (Long.parseLong(a + b) - Long.parseLong(b + a));  // note the overflow of int,max + max
        }
    });
    StringBuilder builder = new StringBuilder();
    for (int i = N - 1; i >= 0; i--)    builder.append(aux[i]);
    int sub = 0;
    for ( ; sub < builder.length() - 1 && builder.charAt(sub) == '0'; sub++);
    return builder.substring(sub).toString();
}

I'm still trying to find ways to avoid using extra space

Solution

The reason is that int and integer are different types Int is a primitive and integer is its object wrapper Comparison < T > applies only to objects; It does not apply to primitives

Put the integer into the container of integers, for example, ArrayList < integer >, and use the comparator to solve the problem

Note that for very large integers, your method may fail because concatenating two valid integers may produce a number too large to hold int You can return to (S1, S2) CompareTo (S2 S1) instead of dictionary comparison, which is the same as that of numbers of the same length