Java数组首尾路径可达性判断方法代码排查求助

Alright, let's walk through the bugs in your code and fix it up properly—your recursive approach is on the right track, but there are a few critical mistakes throwing things off:

Key Issues in Your Original Code #

• Incorrect temp initialization: You created an integer array with int temp = new int [i]; instead of grabbing the step count from the current index. You need int temp = a[i]; to get how far you can move left/right from position i.
• Wrong end condition: The last valid index of an array is a.length - 1, not a.length. Your check if(i==a.length) will never trigger since indices start at 0.
• Overly strict left movement boundary: i-temp>0 blocks you from moving to index 0 (a valid position). It should be i-temp >= 0 to allow movement to the start of the array.
• Order of operations error: You were setting a[i] = 0 (to mark visited) before capturing the step value. You need to save the step count first, otherwise you'll lose the number of steps you can take from that position.

Corrected Code #

public static boolean isWay(int []a){
    // Handle edge cases: null, empty array, or starting with 0 (can't move)
    if(a == null || a.length == 0 || a[0] == 0){
        return false;
    }
    // Clone the array to avoid modifying the original input
    int[] visitedArray = a.clone();
    return isWay(visitedArray, 0);
}

private static boolean isWay(int[] a, int i){
    // Check if we've reached the last index (the array's end)
    if(i == a.length - 1){
        return true;
    }
    // If current position is 0, we've already visited it or can't move
    if(a[i] == 0){
        return false;
    }
    
    // Capture the number of steps we can take from this position
    int stepCount = a[i];
    // Mark current position as visited to prevent infinite loops
    a[i] = 0;
    
    // Check right movement first (if within bounds)
    boolean canMoveRight = (i + stepCount < a.length) && isWay(a, i + stepCount);
    if(canMoveRight){
        return true;
    }
    
    // Check left movement (if within bounds)
    boolean canMoveLeft = (i - stepCount >= 0) && isWay(a, i - stepCount);
    if(canMoveLeft){
        return true;
    }
    
    // Neither direction leads to the end
    return false;
}

Quick Notes on Improvements #

• Cloning the array: This ensures we don't modify the original input array, which is a best practice—callers of your method won't have their array unexpectedly changed.
• Edge case handling: Added checks for null, empty arrays, and starting with 0 to make the method more robust.
• Clear variable names: Renamed temp to stepCount for better readability.

Testing this with your example array [2,4,1,6,4,2,4,3,5] will correctly return true, following the path you described.

内容的提问来源于stack exchange，提问作者Eden Tamam