Leetcode 79. Word Search

题目

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 =[ ['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

分析

在一个字母矩阵中,寻找单词是否存在。使用深度搜索,四个方向搜索,只要找到一个存在即可,由于一个字母不能使用两次,因此深度搜索时候需要先将其赋值为空,之后再恢复即可。

bool search(char** board, int boardRowSize, int boardColSize, char* word,int p,int q,int index)
{
    //printf("%d %d %d\n",p,q,index);
    if(word[index]=='\0')
        return true;
    else
    {
        if(p>0&&word[index]==board[p-1][q])
        {
            char temp=board[p][q];
            board[p][q]='-';
            if(search(board,boardRowSize,boardColSize,word,p-1,q,index+1)==true)
               return true;
            board[p][q]=temp;
        }
        if(q>0&&word[index]==board[p][q-1])
        {
            char temp=board[p][q];
            board[p][q]='-';
            if(search(board,boardRowSize,boardColSize,word,p,q-1,index+1)==true)
               return true;
            board[p][q]=temp;
        }
        if(p<boardRowSize-1&&word[index]==board[p+1][q])
        {
            char temp=board[p][q];
            board[p][q]='-';
            if(search(board,boardRowSize,boardColSize,word,p+1,q,index+1)==true)
               return true;
            board[p][q]=temp;
        }
        if(q<boardColSize-1&&word[index]==board[p][q+1])
        {
            char temp=board[p][q];
            board[p][q]='-';
            if(search(board,boardRowSize,boardColSize,word,p,q+1,index+1)==true)
               return true;
            board[p][q]=temp;
        }
    }
    return false;
}
bool exist(char** board, int boardRowSize, int boardColSize, char* word) {
    bool answer=false;
    for(int i=0;i<boardRowSize;i++)
    {
        for(int j=0;j<boardColSize;j++)
        {
            if(board[i][j]==word[0])
                answer=search(board,boardRowSize,boardColSize,word,i,j,1);
            if(answer==true)return answer;
        }
    }
    return answer;
}
最后编辑于
©著作权归作者所有,转载或内容合作请联系作者
平台声明:文章内容(如有图片或视频亦包括在内)由作者上传并发布,文章内容仅代表作者本人观点,简书系信息发布平台,仅提供信息存储服务。

推荐阅读更多精彩内容