描述

输入：[[a,b,c,e],[s,f,c,s],[a,d,e,e]],"abcced"

返回值：true

示例2

输入：[[a,b,c,e],[s,f,c,s],[a,d,e,e]],"abcb"

返回值：false

备注：

0 <= matrix.length <= 200

0 <= matrix[i].length <= 200

解法思路

遍历矩阵，找到起点，然后使用递归分别从上下左右找到剩余节点，得注意一点就是，当前节点遍历以后，应该替换为特殊字符，防止二次遍历。代码如下

public boolean hasPath (char[][] matrix, String word) {
    // write code here
    if(matrix == null || matrix.length < 1 || word == null){
        return false;
    }
    char[] chars = word.toCharArray();
    for (int i = 0; i < matrix.length; i++) {
        for (int j = 0; j < matrix[0].length; j++) {
             if(hasPathResult(matrix,i,j,chars,0)){
                 return true;
             }
        }
    }
    return false;
}


public boolean hasPathResult(char[][] matrix,int i,int j,char[] word,int index){
    //边界的判断，如果越界直接返回false。index表示的是查找到字符串word的第几个字符，
    //如果这个字符不等于matrix[i][j]，说明验证这个坐标路径是走不通的，直接返回false
    if(i < 0 || i >= matrix.length || j < 0 || j >= matrix[i].length || matrix[i][j] != word[index]){
        return false;
    }
    if(index == word.length-1){
        return true;
    }
    //为了防止当前节点二次使用
    char tmp = matrix[i][j];
    matrix[i][j] = '*';
    boolean flag = hasPathResult(matrix,i+1,j,word,index+1)||hasPathResult(matrix,i,j+1,word,index+1)||
            hasPathResult(matrix,i-1,j,word,index+1) ||hasPathResult(matrix,i,j-1,word,index+1);
    matrix[i][j] = tmp;
    return flag;
}