670. Maximum Swap

Problem Link : https://leetcode.com/problems/maximum-swap/
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Complexity : Time-O(N), Memory-O(N), TimeTaken-2hr
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Logic :
So, whenever we have Qs like to swap a digit of a no, in most of 
the case we have to use stack DS or something simillar.
So here i used stack to store the idx of integer in strictly asc order fashion. 
So the first element of stack will be the smallest and top one will be biggest 
element in stack. After that we will traverse the [] and find the right candidate which satisfy
1. the top element of stack, idx will be > than this element idx
2. The element should also be strictly lesser 
And at any point if any idx satisfies these cond we do swap and we got our answer.


class Solution {
    public int maximumSwap(int num) {
        List<Integer> list=new ArrayList();
        Stack<Integer> st=new Stack();
        while(num>0){
            list.add(num%10);
            num/=10;
        }
        Collections.reverse(list);
        int n=list.size();
        int i=n-1;
        while(i>=0){
            if(st.isEmpty() || list.get(st.peek())<list.get(i)){
                st.push(i);
            } i--;
        }
        
        i=0;
        while(!st.isEmpty() && i<n){
            if(list.get(st.peek())!=list.get(i) && i<st.peek()){
                swap(list, i, st.peek());
                break;
            } else if(i>st.peek()){
                st.pop();
            } else 
                i++;
        }
        
        return toString(list);
    }
    private void swap(List<Integer> list, int i, int j){
        int t1=list.get(i);
        int t2=list.get(j);
        list.set(i, t2);
        list.set(j, t1);
    }
    private int toString(List<Integer> list){
        int num=0;
        for(int i=0;i<list.size();i++){
            num=(num*10)+list.get(i);
        }
        return num;
    }
}