Word Search

/*

Word Search

Given a 2D board and a word, find if the word exists in the grid.
The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.
For example,
Given board =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

*/
class Solution {
public:
    vector<vector<int>> findLetter(vector<vector<char> > board, char c){
     vector<vector<int>> tempResult;
     int i, j;
     int rowSize = board.size();
     if(rowSize==0) return tempResult;
  int colSize = board[0].size();
     if(colSize==0) return tempResult;
     vector<int> onePair;
     for(i=0; i<rowSize; i++){
       for(j=0; j<colSize; j++){
         if(board[i][j]==c){
           onePair.clear();
           onePair.push_back(i);
           onePair.push_back(j);
           tempResult.push_back(onePair);
         }
       }
     }
     return tempResult;
    }
   
        bool containPair(map<vector<int>,bool> used, int row, int col){
  vector<int> pair;
  pair.push_back(row);
  pair.push_back(col);
  if(used.find(pair) != used.end()) return true;
  return false;
 }
   
       void findNextLetter(vector<vector<char>> board, vector<int> temp, int rowIdx, int colIdx, char c, vector<vector<int>> &result){
       vector<int> newLine;
    map<vector<int>, bool> used;
    int i;
    for(i=0; i<temp.size()-1; i=i+2){
     newLine.clear();
     newLine.push_back(temp[i]);
     newLine.push_back(temp[i+1]);
     used[newLine] = true;
    }
       if(colIdx >= 1 && board[rowIdx][colIdx-1]==c && !containPair(used, rowIdx, colIdx-1)) {
                     newLine.clear();
                     newLine = temp;
     newLine.push_back(rowIdx);
     newLine.push_back(colIdx-1);
                     result.push_back(newLine);
    }
       if((colIdx+1)< board[rowIdx].size() && board[rowIdx][colIdx+1]==c && !containPair(used, rowIdx, colIdx+1)) {
                     newLine.clear();
                     newLine = temp;
     newLine.push_back(rowIdx);
     newLine.push_back(colIdx+1);
                     result.push_back(newLine);
    }
       if(rowIdx>=1 && board[rowIdx-1][colIdx]==c && !containPair(used, rowIdx-1, colIdx)) {
                     newLine.clear();
                     newLine = temp;
     newLine.push_back(rowIdx-1);
     newLine.push_back(colIdx);
                     result.push_back(newLine);
    }
       if((rowIdx+1)< board.size() && board[rowIdx+1][colIdx]==c && !containPair(used, rowIdx+1, colIdx)) {
                     newLine.clear();
                     newLine = temp;
     newLine.push_back(rowIdx+1);
     newLine.push_back(colIdx);
                     result.push_back(newLine);
    }
    }
   
    bool exist(vector<vector<char> > &board, string word) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(board.size()==0) return false;
        if(board[0].size()==0) return false;
        int stringSize = word.length();
        if(stringSize==0) return true;
        vector<vector<int>> temp = findLetter(board, word[0]);
        int i = 1;
        int resultSize = temp.size();
        vector<vector<int>>::iterator it;
        vector<int>::iterator lineIt;
        int j;
        int lineSize;
        int rowIdx;
        int colIdx;
  int k;
        while(i<stringSize && resultSize>0){
           for(k=0; k < resultSize && temp[k].size()>=2; k++){
             lineSize = temp[k].size();
             rowIdx = temp[k][lineSize-2];
             colIdx = temp[k][lineSize-1];
             findNextLetter(board, temp[k], rowIdx, colIdx, word[i], temp);
            }
           temp.erase(temp.begin(), temp.begin() + resultSize);
           i++;
           resultSize = temp.size();
        }
       
        for(j=0; j<temp.size(); j++){
          if(temp[j].size() == stringSize*2){
            return true;
          }
        }
       
        return false;
    }
};