如何确定一个字符串矩形的每个边框是否至少包含一个指定字符？

Question

Lets say there is a string that represents a rectangle, and this rectangle only contains two types of chars that are '.' and 'x':

String layout = "..x.\nx..x\n.x.."
System.out.println(layout) ->     ..x.
                                  x..x
                                  .xx.

how can I determine whether each border side of the rectangle has at least one 'x'?(use a boolean method) For example,

//this one is not illegal, because its left side border does not have an 'x' 
 ..x.
 ..xx
 .xx. 

I have put this rectangle in a string[][], which means I have a coordinate for each char in this rectangle. My idea is using four for-loops to check four sides, but it is kind of redundant. Can anyone give a better idea? Thanks

Answer 1

if I understand well, what you want to do than, You can try this. You have any String[][] which holds random Strings and each String must have at least x in it.

 str[0] X . . .
 str[1] . X . .
 str[2] . . X .
 str[3] . . . X

now you need to check str[i] > isContainLatterX() which returns boolean if there is presence of X . For Example :

public boolean isContainLatterX(String str){
    if(str.contains("X"))
          return true;
    else
          return false;
}

Now, need an another Method which returns the Index of occurrence of latter X say indexOfLatterX(String str)

public int indexOfLatterX(String str){
    return str.indexOf("X");
}

Just call both these Methods in Your Loop, and check Latter X is Exists or not, if Not than Generate Error, and skip remaining Progress. if Exists than Check its Index with previous Strings, if there is Matching Index founds than Generate Error and Skip remaining Progress.

Answer 2

Can we try, just getting the co-ordinates in form of [x][y], and check that x and y values exist in range of length... Missing values gives the empty lines

 ..x.
 .x..
 ...x
 x...

The [x][y] list will be (0,2),(1,1),(2,3),(3,0) and must be in range {0,3}

We check that x and y has values (0-3) available. If any value is missing, then the corresponding row or column, don't have an x

Since it represents the border side, you can check for (x,y) in range {0,length-1}