代码随想录三刷day51

最新推荐文章于 2025-12-20 17:21:30 发布

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#深度优先 #算法 #leetcode #动态规划

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前言
一、力扣200. 岛屿数量
二、力扣695. 岛屿的最大面积
三、力扣1020. 飞地的数量
四、力扣130. 被围绕的区域

前言

依然是从地图周边出发，将周边空格相邻的'O'都做上标记，然后在遍历一遍地图，遇到 'O' 且没做过标记的，那么都是地图中间的'O'，全部改成'X'就行。

一、力扣200. 岛屿数量

class Solution {
    int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
    boolean[][] flag;
    public int numIslands(char[][] grid) {
        int m = grid.length, n = grid[0].length;
        int res = 0;
        flag = new boolean[m][n];
        for(int i = 0; i < m; i ++){
            for(int j = 0; j < n; j ++){
                if(!flag[i][j] && grid[i][j] == '1'){
                    res ++;
                    bfs(grid, i, j);
                }
            }
        }
        return res;
    }
    public void bfs(char[][] grid, int x, int y){
        Deque<int[]> deq = new LinkedList<>();
        deq.offerLast(new int[]{x,y});
        while(!deq.isEmpty()){
            int size = deq.size();
            for(int i = 0; i < size; i ++){
                int[] cur = deq.pollFirst();
                for(int j = 0; j < 4; j ++){
                    int curX = cur[0] + arr[j][0];
                    int curY = cur[1] + arr[j][1];
                    if(curX < 0 || curX >= grid.length || curY < 0 || curY >= grid[0].length){
                        continue;
                    }
                    if(!flag[curX][curY] && grid[curX][curY] == '1'){
                        flag[curX][curY] = true;
                        deq.offerLast(new int[]{curX,curY});
                    }
                }
            }
        }
    }
}

二、力扣695. 岛屿的最大面积

class Solution {
    int res = 0;
    int count = 0;
    int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
    boolean[][] flag;
    public int maxAreaOfIsland(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        flag = new boolean[m][n];
        for(int i = 0; i < m; i ++){
            for(int j = 0; j < n; j ++){
                if(!flag[i][j] && grid[i][j] == 1){
                    flag[i][j] = true;
                    count = 1;
                    dfs(grid,i,j);
                }
            }
        }
        return res;
    }
    public void dfs(int[][] grid, int x, int y){
        res = Math.max(count,res);
        for(int i = 0; i < 4; i ++){
            int curX = x + arr[i][0];
            int curY = y + arr[i][1];
            if(curX < 0 || curX >= grid.length || curY < 0 || curY >= grid[0].length){
                continue;
            }
            if(!flag[curX][curY] && grid[curX][curY] == 1){
                flag[curX][curY] = true;
                count ++;
                dfs(grid,curX,curY);
            }
        }
    }
}

三、力扣1020. 飞地的数量

class Solution {
    int res = 0;
    int count = 0;
    int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
    boolean[][] flag;
    boolean f;
    public int numEnclaves(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        flag = new boolean[m][n];
        for(int i = 0; i < m; i ++){
            for(int j = 0; j < n; j ++){
                if(!flag[i][j] && grid[i][j] == 1){
                    flag[i][j] = true;
                    count = 1;
                    f = false;
                    dfs(grid,i,j);
                    if(!f){
                        res += count;
                    }
                }
            }
        }
        return res;
    }
    public void dfs(int[][] grid, int x, int y){
        for(int i = 0; i < 4; i ++){
            int curX = x + arr[i][0];
            int curY = y + arr[i][1];
            if(curX < 0 || curX >= grid.length || curY < 0 || curY >= grid[0].length){
                f = true;
                continue;
            }
            if(!flag[curX][curY] && grid[curX][curY] == 1){
                count ++;
                flag[curX][curY] = true;
                dfs(grid,curX,curY);
            }
        }
    }
}

四、力扣130. 被围绕的区域

class Solution {
    int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
    boolean[][] flag;
    public void solve(char[][] board) {
        int m = board.length, n = board[0].length;
        flag = new boolean[m][n];
        for(int i = 0; i < m; i ++){
            if(!flag[i][0] && board[i][0] == 'O'){
                flag[i][0] = true;
                dfs(board, i, 0);
            }
            if(!flag[i][n-1] && board[i][n-1] == 'O'){
                flag[i][n-1] = true;
                dfs(board,i,n-1);
            }
        }
        for(int i = 0; i < n; i ++){
            if(!flag[0][i] && board[0][i] == 'O'){
                flag[0][i] = true;
                dfs(board,0,i);
            }
            if(!flag[m-1][i] && board[m-1][i] == 'O'){
                flag[m-1][i] = true;
                dfs(board,m-1,i);
            }
        }
        for(int i = 0; i < m; i ++){
            for(int j = 0; j < n; j ++){
                if(!flag[i][j] && board[i][j] == 'O'){
                    board[i][j] = 'X';
                }
            }
        }
    }
    public void dfs(char[][] board, int x, int y){
        for(int i = 0; i < 4; i++){
            int curX = x + arr[i][0];
            int curY = y + arr[i][1];
            if(curX < 0 || curX >= board.length || curY < 0 || curY >= board[0].length){
                continue;
            }
            if(!flag[curX][curY] && board[curX][curY] == 'O'){
                flag[curX][curY] = true;
                dfs(board,curX,curY);
            }
        }
    }
}