My recursive JAVA solution with comment, instead of using 4 for loops to go through each edge

  • 0

    private void spiralOrderHelper(int[][] matrix, int i, int j, int dir, boolean[][] isVisited, ArrayList<Integer> list, int[] di, int[] dj) {

        //if the entry is already visited, no need to add it into list again
        if(!isVisited[i][j]) {
            isVisited[i][j] = true;
        //Check for the end condition for the recursion - the dead node, surrounded by the visited nodes
        //if next entry in current direction has been visited, turn to next direction
        //if next entry in current direction has not been visited, keep going in this direction
        //i+di[dir]: i for the next entry in current direction
        //j+dj[dir]: j for the next entry in current direction
            spiralOrderHelper(matrix, i, j, (dir==3) ? 0 : dir+1, isVisited, list, di, dj);
            spiralOrderHelper(matrix, i+di[dir], j+dj[dir], dir, isVisited, list, di, dj);
    public List<Integer> spiralOrder(int[][] matrix) {
        if(matrix==null || matrix.length==0)    return new ArrayList<Integer>();
        int[] di = {0, 1, 0, -1};
        int[] dj = {1, 0, -1, 0};
        ArrayList<Integer> resultList = new ArrayList<Integer>();
        boolean[][] isVisited = new boolean [matrix.length+2][matrix[0].length+2];
        for(int i=0; i<isVisited.length; i++) {
            isVisited[i][0] = isVisited[i][isVisited[0].length-1] = true;
        for(int j=0; j<isVisited[0].length; j++) {
            isVisited[0][j] = isVisited[isVisited.length-1][j] = true;
        //Dirction: 0: Right    1: Down    2: Left    3: Up
        spiralOrderHelper(matrix, 1, 1, 0, isVisited, resultList, di, dj);
        return resultList;

  • 0

    Good for practice, but O(n*m) space + an O(n*m) deep recursion can't compete with an O(1) space and O(n*m) time simple iterative algorithm.

    You can save stack space by putting di and dj as private static final, because they don't change, ever, so it's wasteful to pass the around.

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