第四章 Django 模板系统

文章目录

• 题目描述
• 思路
• 解题方法
• 复杂度
• Code

题目描述

思路

该问题可以归纳为一类遍历二维矩阵的题目，此类中的一部分题目可以利用DFS来解决，具体到本题目：

解题方法

复杂度

时间复杂度:

空间复杂度:

Code

class Solution {
    private boolean[][] visited;
    private int row;
    private int col;

    /**
     * Get all the island counts
     *
     * @param grid Given a two-dimensional array
     * @return int
     */
    public int numIslands(char[][] grid) {
        row = grid.length;
        col = grid[0].length;
        visited = new boolean[row][col];
        //The count of island
        int count = 0;
        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < col; ++j) {
                if (visited[i][j] != true && grid[i][j] == '1') {
                    count++;
                    dfs(grid, i, j);
                }
            }
        }
        return count;
    }

    /**
     * Try dfs or not from each point in a two-dimensional array
     *
     * @param grid Given a two-dimensional array
     * @param i    Abscissa
     * @param j    Ordinate
     */
    private void dfs(char[][] grid, int i, int j) {
        //Record four bearings
        int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        //The current legal location is set to true
        visited[i][j] = true;
        for (int k = 0; k < 4; ++k) {
            int newI = i + directions[k][0];
            int newJ = j + directions[k][1];
            if (newI >= 0 && newI < row && newJ >= 0 && newJ < col
                    && visited[newI][newJ] == false && grid[newI][newJ] == '1') {
                dfs(grid, newI, newJ);
            }
        }
    }
}