Java | 71% faster and 21 % faster solutions
Ex: 1 - Lets n is 10 , and start = rounds[0] is 5 and end i.e rounds[rounds.length-1] is 10\nthe more number of visted set/number will be 5,6,7,8,9,10 as those 
Ex: 1 - Lets n is 10 , and start = rounds[0] is 5 and end i.e rounds[rounds.length-1] is 10\nthe more number of visted set/number will be 5,6,7,8,9,10 as those are going to be common in cicular condition. as you can\'t by pass them to go to other set/number. As your start and end are fixed how many round you will do these numbers are going to remain max visited set.\n\nEx: 2 - Lets n is 10 , and start = rounds[0] is 5 and end i.e rounds[rounds.length-1] is 2\nthe more number of visted set/number will be 5,6,7,8,9,10,1,2 as those are going to be common in cicular condition. as you can\'t by pass them to go to other set/number.\n\nby putting these two condition you can figure out the number.\n\nNow the result need in ascending order. \nFor the 2nd example it is tricky so you have to handel in two loop. starting from `1` to `end`, and `start` to `n`   \nfor first example it is one loop with `start` to `end`\n \n\n```\n//71% faster\n List<Integer> ans = new ArrayList<>();\n        int st = rounds[0];\n        int end = rounds[rounds.length-1];\n        if(st <= end){\n            while(st<=end){\n                ans.add(st++);\n            }\n        }\n        else{\n            int x =1;\n            while(x<=end){\n                ans.add(x++);\n            }\n            while(st<=n){\n                ans.add(st++);\n            }\n        }\n        \n        return ans;\n```\n\nAfter Finding Number doing collections sort increases the time complexity \n```\n//21% faster\npublic List<Integer> mostVisited(int n, int[] rounds) {\n        List<Integer> ans = new ArrayList<>();\n        int st = rounds[0]; \n        while (st != rounds[rounds.length-1]){\n            ans.add(st);\n            st++;\n            if(st>n) st=1;\n        }\n        ans.add(st);\n        Collections.sort(ans);\n        return ans;\n    }\n```
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