Change array in-place

https://discuss.leetcode.com/topic/98/change-array-in-place
Given an array of length n having integers 0 to n-1 in unsorted order.
Please modify this array such that the value at a[i] becomes a[a[i]].
For example, if a[0] = 5, a[0] will have value a[5] and so on.
e.g. {2, 4, 3, 1, 0} => {3, 0, 1, 4, 2}
This should take O(n) time complexity

 I think that here the trick is that every permuatation is represented as a product of disjoint cycles, and we have to follow all cycles. When we finish one cycle, I suggest to mark numbers from the cycle in some way (put them negative) to avoid visiting the same cycle twice In the example above we have only one cycle (0-2-3-1-4-0)

Method 1, visited numbers in the cycle are set negative

void changeInPlace(int[] nums) {
         int maxVal = Integer.MIN_VALUE;  
  for (int i = 0; i < nums.length; i++) {
        if (nums[i] > 0) {
     int startIndex = i;
     int nextIndex = nums[startIndex];
     while (i != nextIndex) {     
   int next = nums[nextIndex];
   swap(nextIndex,startIndex, nums);
          nums[startIndex]+=maxVal;
   startIndex = nextIndex;
    nextIndex = next;      
      
        }
  nums[startIndex]+=maxVal;   
}
}
for (int i = 0;i < nums.length; i++) {
 System.out.println(nums[i]-maxVal);
 }
}

Another test which is very illustrative to observe multiple cycles is : {6,4,2,3,1,5,0}
Time complexity O(n)
Method 2, visited numbers in the cycle are increased with nums.length
Here is an more triecker way :

void changeInPlace(int[] nums) {
 for (int i = 0; i < nums.length; i++) {
         if (nums[i] < nums.length) {
       int startIndex = i;
       int nextIndex = nums[startIndex];
        while (i != nextIndex) {     
    int next = nums[nextIndex];
    swap(nextIndex,startIndex, nums);
    nums[startIndex]+=nums.length;
    startIndex = nextIndex;
    nextIndex = next;      
      
    }
   nums[startIndex]+=nums.length;   
  }
    
 }
 for (int i = 0;i < nums.length; i++) {
       System.out.println(nums[i]-nums.length);
 }