public void printNum(int[] arr) {
int[] result = new int[2];
int[] heap = new int[101];
for (int i = 0; i < 98; i++) {
heap[arr[i]] = 1;
}
for (int i = 1; i <= 100; i++) {
if (heap[i] != 1) {
System.out.println(i);
}
}
}
public int[] findNum(int[] arr) {
int[] result = new int[2];
var count = 0;
for (int i = 1; i <= 100; i++) {
if (count == 2) break;
if (arr.indexOf(i) == -1) {
result.push(i);
count++;
}
}
return result;
}
有人给出了一个时间复杂度O(n^2)的算法,我给一个O(n)的算法把: