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  • LeetCode-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:
        bool Sub(vector<vector<char> >&board,string& word,int index,int i,int j,vector<vector<bool> > &used){
            if(index==word.length())return true;
            if(i>=1&&!used[i-1][j]){
                used[i-1][j]=true;
                if(board[i-1][j]==word[index]&&Sub(board,word,index+1,i-1,j,used))return true;
                used[i-1][j]=false;
            }
            if(i<board.size()-1&&!used[i+1][j]){
                used[i+1][j]=true;
                if(board[i+1][j]==word[index]&&Sub(board,word,index+1,i+1,j,used))return true;
                used[i+1][j]=false;
            }
            if(j>=1&&!used[i][j-1]){
                used[i][j-1]=true;
                if(board[i][j-1]==word[index]&&Sub(board,word,index+1,i,j-1,used))return true;
                used[i][j-1]=false;
            }
            if(j<board[0].size()-1&&!used[i][j+1]){
                used[i][j+1]=true;
                if(board[i][j+1]==word[index]&&Sub(board,word,index+1,i,j+1,used))return true;
                used[i][j+1]=false;
            }
            return false;
        }
        bool exist(vector<vector<char> > &board, string word) {
            // Note: The Solution object is instantiated only once and is reused by each test case.
            if(word=="")return false;
            if(board.size()==0||board[0].size()==0)return false;
            vector<vector<bool> > used;
            vector<bool> one;
            one.resize(board[0].size(),false);
            used.resize(board.size(),one);
            for(int i=0;i<board.size();i++){
                for(int j=0;j<board[i].size();j++){
                    used[i][j]=true;
                    if(board[i][j]==word[0]&&Sub(board,word,1,i,j,used))return true;
                    used[i][j]=false;
                }
            }
            return false;
        }
    };
    View Code
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  • 原文地址:https://www.cnblogs.com/superzrx/p/3349490.html
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