如果在java語句行爲奇怪

Question

我在java中有一個奇怪的問題與if-else語句。下面是一個遞歸方法，試圖找到一個名爲getPathThroughMaze的迷宮的末尾。

private static String getPathThroughMaze(char[][] maze, Set<Point> visited, Point currentPoint) { 
    int currentX = currentPoint.x; 
    int currentY = currentPoint.y; 
    visited.add(currentPoint); 

    //end case. append '!' so we know which path leads to the end of the maze 
    if (currentX == (xLength - 2) && currentY == (yLength - 1)) { 
     return "!"; 
    } 

    char left = maze[currentY][currentX - 1]; 
    char right = maze[currentY][currentX + 1]; 
    char up = maze[currentY - 1][currentX]; 
    char down = maze[currentY + 1][currentX]; 
    char current = maze[currentY][currentX]; 

    /* If valid, visit all non-visited adjacent squares. 
     Only odd numbered columns will be traversed 
     since even numbered columns represent vertical wall 
     columns 
    */ 
    if (right == '_' || right == ' ') { 
     Point nextPoint = new Point(currentX + 2, currentY); 
     if (!visited.contains(nextPoint)) { 
      String path = "E" + getPathThroughMaze(maze, visited, nextPoint); 
      if (path.endsWith("!")) { 
       return path; 
      } else { 
       //do nothing. 
      } 
     }else { 
      //do nothing. 
     } 
    } else if (up == ' ') { 
     Point nextPoint = new Point(currentX, currentY - 1); 
     if (!visited.contains(nextPoint)) { 
      String path = "N" + getPathThroughMaze(maze, visited, nextPoint); 
      if (path.endsWith("!")) { 
       return path; 
      } else { 
       //do nothing. 
      } 
     } else { 
      //do nothing. 
     } 
    } else if (current == ' ' && (down == '_' || down == ' ')) { 
     Point nextPoint = new Point(currentX, currentY + 1); 
     if (!visited.contains(nextPoint)) { 
      String path = "S" + getPathThroughMaze(maze, visited, nextPoint); 
      if (path.endsWith("!")) { 
       return path; 
      } else { 
       //do nothing. 
      } 
     } else { 
      //do nothing. 
     } 
    } else if (left == '_' || left == ' ') { 
     Point nextPoint = new Point(currentX - 2, currentY); 
     if (!visited.contains(nextPoint)) { 
      String path = "W" + getPathThroughMaze(maze, visited, nextPoint); 
      if (path.endsWith("!")) { 
       return path; 
      } else { 
       //do nothing. 
      } 
     } else { 
      //do nothing. 
     } 
    } else { 
     return ""; 
    } 
    //otherwise... 
    return ""; 
} 

在遞歸在那裏我遇到一個問題點的變量是：

currentX = 3 
currentY = 2 
right = '|' 
left = '|' 
up = ' ' 
down = '_' 
current = ' ' 
visited contains points (1,1), (3,1), (3,2) 

在第一個else if語句：

else if (up == ' ') 

一個新的點（3， 1）被創建，其已被包含在訪問集中。我希望發生的是，

if(!visited.contains(nextPoint)) 

將評估爲假，我會（後也許在調試器上點擊幾步驟）在

else if (current == ' ' && (down == '_' || down == ' ')) 

到哪我就可以檢查條件（我認爲這是真的），並繼續穿越迷宮。而實際上，當我點擊跨過上

if(!visited.contains(nextPoint)) 

調試器（在elcipse和IntelliJ）移動一直到我的方法的最後一個return語句，並想要返回「」。我不明白爲什麼我所有其他的陳述都被跳過了。任何人都可以啓發我，爲什麼可能是這種情況？如果我的解釋不夠清楚，請讓我知道。

Answer 1

If/else陳述是排他性的，因此您不會到達else if (current == ' ' && (down == '_' || down == ' '))，因爲您已經進入else if (up == ' ')分支。由於if(!visited.contains(nextPoint))內部的if爲false，程序進入else部分，//do nothing註釋並且什麼都不做（實際上，你不需要也不應該寫一個空的else語句，至少要添加一個log語句給它使其更易於調試）。然後它退出if/else區塊並前往return。

如果您希望您的代碼在每次方法調用時檢查每個分支的if/else，只需將其替換爲一些簡單的if語句即可。

I.e.而不是：

if (condition1){ 
} else if (condition2){ 
} else if (condition3){ 
} 

寫

if (condition1){ 
} 
if (condition2){ 
} 
if (condition3){ 
}