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前言

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岛屿问题最容易让人想到BFS或者DFS，但是这道题还真的没有必要，别把简单问题搞复杂了

一、力扣417. 太平洋大西洋水流问题

class Solution {int[][] arr = new int[][]{{0,1},{0,-1},{-1,0},{1,0}};boolean[][][] flag;public List<List<Integer>> pacificAtlantic(int[][] heights) {int m = heights.length, n = heights[0].length;flag = new boolean[m][n][2];List<List<Integer>> res = new ArrayList<>();for(int i = 0; i < m; i ++){if(!flag[i][0][0]){dfs(heights,i,0,0);}if(!flag[i][n-1][1]){dfs(heights,i,n-1,1);}}for(int i = 0; i < n; i ++){if(!flag[0][i][0]){dfs(heights,0,i,0);}if(!flag[m-1][i][1]){dfs(heights,m-1,i,1);}}for(int i = 0; i < m; i ++){for(int j = 0; j < n; j ++){if(flag[i][j][0] && flag[i][j][1]){res.add(new ArrayList<>(Arrays.asList(i,j)));}}}return res;}public void dfs(int[][] heights, int x, int y, int sign){flag[x][y][sign] = true;for(int i = 0; i < 4; i ++){int curX = x + arr[i][0];int curY = y + arr[i][1];if(curX < 0 || curX >= heights.length || curY <0 || curY >= heights[0].length){continue;}if(!flag[curX][curY][sign] && heights[curX][curY] >= heights[x][y]){dfs(heights,curX,curY,sign);}}}
}

二、力扣827. 最大人工岛

class Solution {int[][] arr = new int[][]{{1,0},{-1,0},{0,1},{0,-1}};int count;public int largestIsland(int[][] grid) {Map<Integer,Integer> map = new HashMap<>();int flag = 2;int res = Integer.MIN_VALUE;int m = grid.length, n = grid[0].length;for(int i = 0; i < m; i ++){for(int j = 0; j < n; j ++){if(grid[i][j] == 1){count = 1;grid[i][j] = flag;dfs(grid,i,j,flag);map.put(flag,count);flag ++;}}}for(int i = 0; i < m; i ++){for(int j = 0; j < n; j ++){if(grid[i][j] == 0){Set<Integer> set = new HashSet<>();int sum = 0;for(int k = 0; k < 4; k ++){int curX = i + arr[k][0];int curY = j + arr[k][1];if(curX < 0 || curX >= m || curY < 0 || curY >= n){continue;}if(set.contains(grid[curX][curY]) || !map.containsKey(grid[curX][curY])){continue;}sum += map.get(grid[curX][curY]);set.add(grid[curX][curY]);}res = Math.max(res,sum);}}}return res == Integer.MIN_VALUE ? m * n : res+1;}public void dfs(int[][] grid, int x, int y,int flag){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(grid[curX][curY] == 1){count ++;grid[curX][curY] = flag;dfs(grid,curX,curY,flag);}}}
}

三、力扣127. 单词接龙

class Solution {public int ladderLength(String beginWord, String endWord, List<String> wordList) {Deque<String> deq = new LinkedList<>();int res = 1;Map<String,Boolean> map = new HashMap<>();for(String s : wordList){map.put(s,false);}if(wordList.size() == 0 || !map.containsKey(endWord)){return 0;}deq.offerLast(beginWord);while(!deq.isEmpty()){int size = deq.size();for(int j = 0; j < size; j ++){String cur = deq.pollFirst();for(int i = 0; i < wordList.size(); i ++){String next = wordList.get(i);if(!map.get(next) && fun(cur,next)){if(next.equals(endWord)){return res+1;}map.put(next,true);deq.offerLast(next);}}}res ++;}return 0;}public boolean fun(String cur, String next){int dif = 0;for(int i = 0; i < cur.length(); i ++){if(cur.charAt(i) != next.charAt(i)){dif ++;}if(dif >= 2){return false;}}if(dif == 0){return false;}return true;}
}

四、力扣841. 钥匙和房间

class Solution {public boolean canVisitAllRooms(List<List<Integer>> rooms) {Map<Integer,Boolean> map = new HashMap<>();for(int i = 0; i < rooms.size(); i ++){map.put(i,false);}map.put(0,true);Deque<Integer> deq = new LinkedList<>();deq.offerLast(0);while(!deq.isEmpty()){int size = deq.size();for(int i = 0; i < size; i++){int cur = deq.pollFirst();for(int j = 0; j < rooms.get(cur).size(); j ++){if(!map.get(rooms.get(cur).get(j))){deq.offerLast(rooms.get(cur).get(j));map.put(rooms.get(cur).get(j),true);}}}}for(Map.Entry<Integer,Boolean> entrys: map.entrySet()){if(!entrys.getValue()){return false;}}return true;}
}