838. Push Dominoes --M

 There are N dominoes in a line, and we place each domino vertically upright.

In the beginning, we simultaneously push some of the dominoes either to the left or to the right.

After each second, each domino that is falling to the left pushes the adjacent domino on the left.

Similarly, the dominoes falling to the right push their adjacent dominoes standing on the right.

When a vertical domino has dominoes falling on it from both sides, it stays still due to the balance of the forces.

For the purposes of this question, we will consider that a falling domino expends no additional force to a falling or already fallen domino.

Given a string "S" representing the initial state. S[i] = 'L', if the i-th domino has been pushed to the left; S[i] = 'R', if the i-th domino has been pushed to the right; S[i] = '.', if the i-th domino has not been pushed.

Return a string representing the final state. 

Example 1:

Input: ".L.R...LR..L.."
Output: "LL.RR.LLRRLL.."

Example 2:

Input: "RR.L"
Output: "RR.L"
Explanation: The first domino expends no additional force on the second domino.

Note:

1. 0 <= N <= 10^5
2. String dominoes contains only 'L', 'R' and '.'

A:

--------BFS---------  

class Solution {
public:
    string pushDominoes(string dominoes) {
        vector<char> V(dominoes.begin(), dominoes.end());
        int n = V.size();
        vector<vector<int>> curList; // each <index, direction(-1/1) >, -1 is L, 1 is right
        for(int i =0;i< n;i++){
            if(V[i] == 'L')
                curList.push_back(vector<int>{i,-1});
            if(V[i] == 'R')
                curList.push_back(vector<int>{i,1});
        }
        vector<int> curPos(n,0); // record current position would go left or right, or both
        while(not curList.empty()){            
            vector<vector<int>> nextList;
            for(auto val : curList){
                int index = val[0], direction = val[1];
                if(direction + index >=0 && direction +index <n && V[index+direction] == '.'){
                    curPos[index+direction]+=direction;
                    nextList.push_back(vector<int>{index + direction, direction});
                }
            }
            for(int i =0;i<V.size();i++){
                if(curPos[i] == -1)
                    V[i] = 'L';                
                if(curPos[i] == 1)
                    V[i] = 'R';
                curPos[i] = 0;
            }            
            curList = nextList;
        }
        string res(V.begin(), V.end());
        return res;
    }
};

Miestakes：

第一遍的时候，没有考虑左右都倒过来的情况。这样是不变的。