【leetcode】838. Push domino (Analog)

subject ：
838. Push domino
n A row of dominoes , Erect each domino vertically . At the beginning , At the same time, push some dominoes to the left or right .

Every second , The dominoes falling to the left will push the dominoes adjacent to the left . similarly , The dominoes on the right will also push the adjacent dominoes on the right .

If a vertically erected dominoes is falling on both sides at the same time , Because of the balance of forces , The dominoes remain the same .

As far as this is concerned , We would assume that a falling domino does not exert additional force on other falling or fallen dominoes .

Give you a string dominoes Indicates the initial state of this row of dominoes , among ：

dominoes[i] = ‘L’, It means the first one i A domino is pushed to the left ,
dominoes[i] = ‘R’, It means the first one i A domino is pushed to the right ,
dominoes[i] = ‘.’, It means that the third party has not been promoted i Dominoes .
Returns a string representing the final state .

Example 1：

Input ：dominoes = “RR.L”
Output ：“RR.L”
explain ： The first dominoes didn't put extra force on the second one .
Example 2：

Input ：dominoes = “.L.R…LR…L…”
Output ：“LL.RR.LLRRLL…”

Tips ：

n == dominoes.length
1 <= n <= 105
dominoes[i] by ‘L’、‘R’ or ‘.’

simulation ：

We can enumerate all consecutive dominoes that have not been pushed , According to the two sides of this piece of dominoes （ If any ） The direction of pushing down determines the final state of this dominoes ：

• If the dominoes on both sides are in the same direction , Then this continuous vertical dominoes will fall in the same direction .
• If the dominoes on both sides are opposite , Then this piece of dominoes will fall to the middle .
• If the dominoes on both sides are opposite , Then the dominoes will remain upright .
  Specially , If there is no dominoes pushed down on the left , Then suppose there is a dominoes that fall to the left ; If there are no dominoes pushed down on the right , Then suppose there is a dominoes that fall to the right . This assumption will not destroy the final state of the dominoes , And the boundary case can also fall into the above three cases .

We can use two pointers i and j Not all dominoes are pushed continuously ,left and right Indicates the direction of the dominoes on both sides . Calculate the final state of the dominoes according to the above three cases .

class Solution {
    
    //  simulation 
    public String pushDominoes(String dominoes) {
    
        char[] s = dominoes.toCharArray();
        int n = s.length, i = 0;
        char left = 'L';
        while (i < n) {
    
            int j = i;
            while (j < n && s[j] == '.') {
     //  Find a continuous piece of dominoes that have not been pushed 
                j++;
            }
            char right = j < n ? s[j] : 'R';
            if (left == right) {
     //  In the same direction , Then these erected dominoes will fall in the same direction 
                while (i < j) {
    
                    s[i++] = right;
                }
            } else if (left == 'R' && right == 'L') {
     //  Relative direction , Then fall from both sides to the middle 
                int k = j - 1;
                while (i < k) {
    
                    s[i++] = 'R';
                    s[k--] = 'L';
                }
            }
            left = right;
            i = j + 1;
        }
        return new String(s);
    }
}