BFS

1091. 二进制矩阵中的最短路径

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思路：BFS
从起点出发，往8个方向（左，右，上，下，左上，左下，右上，右下）分别遍历，也就是考虑每种情况，然后不断向外扩，而且仅考虑为0的节点。由于求路径长度，所以需要知道当节点到达最后一个节点时，是第几层的BFS，于是构造了Node节点，用于存放在矩阵中的位置和BFS层数。

提示：如果不是很懂的话，建议学下BFS，特别是里面的迷宫例子，流程基本很像。这道题就是判断比较多，不然代码看着也不会那么吓人，其实基本步骤能写出来就行，后面那些判断再一点点加上去就行了。

class Solution {//进阶：用grid数组代替inq数组，也就是走过的标为1就行了。//bfspublic int shortestPathBinaryMatrix(int[][] grid) {//队列和第一个元素初始化Queue<Node> queue = new LinkedList<>();        if(grid[0][0] == 0) queue.add(new Node(0,0,1));int len = grid[0].length;boolean [][] inq = new boolean[len][len];//记录是否访问，防止走回头路inq[0][0] = true;if(len == 1) return 1;int result = -1;//结果while(!queue.isEmpty()) {Node tmp = queue.poll();int x = tmp.getX();int y = tmp.getY();//遍历8个方向上的位置//x-1,y-1  x-1,y  x-1,y+1  x,y-1  x,y(本身)  x,y+1  x+1,y-1  x+1,y  x+1,y+1for(int i=x-1; i<=x+1; i++) {      if(i<0 || i>=len) continue;//边界for(int j=y-1; j<=y+1; j++) {if(j<0 || j>=len) continue;//边界if(i==x && y==j) continue;//本身//未访问且为0if(!inq[i][j] && grid[i][j]==0) {queue.add(new Node(i,j,tmp.getLayer()+1));inq[i][j] = true;}}}                              if(inq[len-1][len-1]) {result = tmp.getLayer()+1; //因为tmp是当前层           break;}}return result;}
}
//保存每个节点
class Node{private int x;//行下标private int y;//列下标private int layer;//层次public Node(int x, int y, int layer) {this.x = x;this.y = y;this.layer = layer;}public int getX() {return x;}public int getY() {return y;}public int getLayer() {return layer;}
}

279. 完全平方数

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思路：BFS
计算n范围内的完全平方数，例如12的为[1,4,9]，从12出发可以有三条路可走，计算每条路的剩余值为[11,8,3]（加入队列，作为下一轮）; 以每个剩余值为下一个起点，计算它们能走的路，例如11的可选完全平方数为[1,4,9], 8的为[1,4], 3的为[1]，计算走每条路的剩余值，以此类推，直到剩余为0时，返回当前层数。BFS的特点是找到后就结束，比暴力for循环优。

class Solution {public int numSquares(int n) {//保存完全平方数List<Integer> ps = getPerfectSquare(n);//保存剩余值，一开始是nQueue<Integer> queue = new LinkedList<>();         queue.add(n);//层数int layer = 0;while(!queue.isEmpty()) {layer++;            //遍历每层int size = queue.size();while(size-->0){Integer rest = queue.poll();                //计算能走的路for(int i=ps.size() - 1 ; i>=0; i--) {  //从大到小，节省时间//跳过大于剩余值的元素if(ps.get(i) <= rest) {int tmp = rest - ps.get(i);if(tmp == 0) return layer;queue.add(tmp);}}}}return 0;}//利用加法求完全平方数1,4,9,16...相差3,5,7(奇数)，加快求解速度List<Integer> getPerfectSquare(int n) {List<Integer> ps = new ArrayList<>();int num = 1;int i = 3;//代表奇数        while(n >= num) {ps.add(num);num = num + i;i+=2;}return ps;}
}

优化：在上面的图中，我们会发现重复遍历的情况，像1，9和1，9其实是一样的，包括4，1和1，4，虽然不影响结果，但影响效率。
具体解决有两种，一种是遵循一定规则，例如拆分时，每次拆分的元素不大于上一个元素，也就是选择的平方数的时候，不能大于前面的选择，这就需要记录上一个平方数的值了，我们可以构造一个类，包含上一次选中的平方数和剩余值）

另一种方式就是使用访问数组，具体参考下一题，下面仅以遵循一定规则为例

class Solution {public int numSquares(int n) {List<Integer> ps = getPerfectSquare(n);//保存剩余值和上一个平方数，一开始都是nQueue<Record> queue = new LinkedList<>(); queue.add(new Record(n,n));int layer = 0;        while(!queue.isEmpty()) {layer++;            int size = queue.size();while(size-->0){Record rd = queue.poll();for(int i=ps.size() - 1 ; i>=0; i--) {                    //下一个平方数必须不能大于上一个平方数，且不能大于剩余值if(ps.get(i) <= rd.prev && ps.get(i) <= rd.rest) {int tmp = rd.rest - ps.get(i);if(tmp == 0) return layer;queue.add(new Record(ps.get(i),tmp));}}}}return 0;}List<Integer> getPerfectSquare(int n) {List<Integer> ps = new ArrayList<>();int num = 1;int i = 3;//代表奇数        while(n >= num) {ps.add(num);num = num + i;i+=2;}return ps;}
}class Record{    int prev;//上一个平方数int rest;//剩余值public Record(int prev, int rest){this.prev = prev;this.rest = rest;}
}

322. 零钱兑换

方法一：BFS，这里采用的是使用访问数组，如果使用遵循一定规则，会出现超时情况。
优化：我们需要转换思路，如果按完全平方数那种做法，例如[1,2,5] 100，这个用例BFS深度就达到20层，而且随着深度越深，队列就越长。原因就出在我所构造的路径节点是以数组中的值来决定的，而这些值（1，2，5）与目标值100相差甚远，而不像完全平方数，数据的选择是100以内的完全平方数。所以这道题需要转变思路：路径节点由剩余值来决定，而不是数组中元素的值，例如上面的例子，节点的取值为100，减去[1，2，5]这些元素，剩下的值作为路径节点，然后使用访问数组标记访问过的无需访问（100-2 和100-1-1是一样的，都是98)，也就是整个BFS的节点个数不会超过100个，具体对比如下图(只画了5层)，第一张图改变了节点的选择，并使用了访问数组，到了第五层，节点个数基本就开始固定了，即便遍历完整个图，也只需要100个节点。但对于采用数组元素作为节点，其每层的节点个数在不断增加。

class Solution {public int coinChange(int[] coins, int amount) {//特殊情况直接返回if(amount == 0)  return 0;//先排序，这样就能先遍历大的数---不排序也okArrays.sort(coins);        //标记数组boolean[] visited = new boolean[amount];Queue<Integer> queue = new LinkedList<>();queue.add(amount);        //层数，也代表最少硬币数int layer = 0; while(!queue.isEmpty()) {layer++;int size = queue.size();while(size-- > 0) {int rest = queue.poll();   //从大到小                                     for(int i = coins.length - 1; i >= 0; i--) {         //不能大于剩余值if(coins[i] <= rest) {int tmp = rest - coins[i];if(tmp == 0) return layer;//未访问过才能加入队列if(!visited[tmp]) {                            queue.add(tmp);visited[tmp] = true;}}                                    }}}return -1;                }
}

DFS

200. 岛屿数量

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思路一：DFS
遇到为1的点，就以该点为起点，往四个方向搜索（上下左右），每一次搜索完毕，代表一个岛屿构建完毕。对同一个岛的点，在一次遍历后无需再次遍历，因为只要是在同一个岛上，无论从那个点开始遍历，遍历完之后还是那个岛，所以需要对同一个岛上的点标记是否访问过，避免重复遍历（而不是说遇到1就遍历一次，那岛的数量就是1的个数了，显然不对）。
具体如何标记，这里可以开一个布尔类型的数组，但仔细观察，矩阵中为0的点我们是鸟都不用鸟的，说明只要将访问过的点置为零就行了，也就是grid[x][y]=‘0’。

class Solution {private int m;//行private int n;//列public int numIslands(char[][] grid) {m = grid.length;n = grid[0].length;int count = 0;for(int i = 0; i < m; i++) {for(int j = 0; j < n; j++) {if(grid[i][j] == '1') {count++;//每一次搜索相当于构建了一个岛屿dfs(grid, i, j);}                }}return count;}//往四个方向深度搜索//目的是为了将同一个岛屿上1的点标记为0，代表访问过。防止同一个岛屿重复访问。public void dfs(char[][] grid, int x, int y) {if(x < 0 || y < 0 || x >=m || y >= n || grid[x][y]=='0') return;//边界grid[x][y] = '0';//标记访问过dfs(grid, x-1, y);dfs(grid, x+1, y);dfs(grid, x, y-1);dfs(grid, x, y+1);}
}

思路二：BFS
类似DFS，只是把递归改为队列迭代而已

//存放点的位置
class Node{int x;int y;public Node(int x, int y){this.x = x;this.y = y;}
}
class Solution {private int m;private int n;public int numIslands(char[][] grid) {m = grid.length;n = grid[0].length;int count = 0;for(int i = 0; i < m; i++) {for(int j = 0; j < n; j++) {if(grid[i][j] == '1'){                    count++;//每一次搜索相当于构建了一个岛屿bfs(grid, i, j); }                                               }}return count;}final int[][] XY = {{-1,0},{1,0},{0,-1},{0,1}};//往四个方向发散搜索public void bfs(char[][] grid, int i, int j) {        grid[i][j] = '0';//标记访问过Queue<Node> fifo = new LinkedList<>();fifo.add(new Node(i, j));while(!fifo.isEmpty()) {Node node = fifo.poll();for(int loop = 0; loop < 4; loop++) {int newx = node.x + XY[loop][0];int newy = node.y + XY[loop][1];if(!judge(newx, newy) && grid[newx][newy] == '1') {fifo.add(new Node(newx, newy));grid[newx][newy] = '0';//标记访问过}}   }       }//边界判断boolean judge(int x, int y) {return x<0 || x>=m || y<0 || y>=n;}
}

695. 岛屿的最大面积

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思路一：DFS
岛屿的数量求的是矩阵中有多少个岛屿，而本题求的是所有岛屿中，那个岛屿的面积最大，即占有1的点数谁最多，思路基本跟求岛屿的数量一样，唯一不同的是，对每个岛的遍历，需要进行计数，且每遍历一个岛屿就和最大值进行比较，并归零进入下个岛屿计数。可根据上题代码进行修改，这里给出其它不同写法，但思路都一样。

class Solution {private int max = 0;//最大计数private final int[] direX= {-1,1,0,0};//往四个方向遍历private final int[] direY= {0,0,-1,1};private int m,n;//行数、列数private int count;//计数public int maxAreaOfIsland(int[][] grid) {m = grid.length;//行n = grid[0].length;//列for(int i=0; i<m; i++) {for(int j=0; j<n; j++) {                if(grid[i][j]==0) continue;count=0;//归零，遍历每个岛都要从0开始                        dfs(grid,i,j);                if(count > max) max = count; //最大值记录               }}return max;}//只有符合的点才能进入dfspublic void dfs(int[][] grid, int x, int y) {                        grid[x][y]=0;//访问过置零count++;//累加int four = 0;while(four<4) {int idx = x + direX[four];int idy = y + direY[four];            four++;if(idx<0 || idy<0 || idx>=m || idy>=n || grid[idx][idy] == 0) continue;    dfs(grid,idx,idy);                                                       }        }
}

思路二：BFS

class Solution {private int max = 0;//最大计数private final int[] direX = {-1, 1, 0, 0};//往四个方向遍历private final int[] direY = {0, 0, -1, 1};int m, n;public int maxAreaOfIsland(int[][] grid) {m = grid.length;//行n = grid[0].length;//列for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (grid[i][j] == 0) continue;bfs(grid, i, j);}}return max;}public void bfs(int[][] grid, int x, int y) {//队列Queue<Node> qe = new LinkedList<>();qe.add(new Node(x, y));grid[x][y] = 0;//首个int count = 1;//计数while (!qe.isEmpty()) {Node node = qe.poll();int four = 0;while (four < 4) {int dx = node.x + direX[four];int dy = node.y + direY[four];four++;//四个方向遍历if (dx < 0 || dx >= m || dy < 0 || dy >= n || grid[dx][dy] == 0) continue;qe.add(new Node(dx, dy));grid[dx][dy] = 0;count++;}}if (count > max) max = count;//更新最大值}
}//存放点的类
class Node {int x;int y;Node(int x, int y) {this.x = x;this.y = y;}
}

547. 省份数量

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思路一：DFS
该题和岛屿数量类似，但并不一样，岛屿数量通过矩阵可肉眼看出有多少个，但这里并不能，因为这里的点代表两个城市的关系（坐标x,y代表两个城市），而岛屿数量是点与点之间的关系（所以能肉眼看出）。
遍历的时候需要按城市之间的关系去遍历，例如遍历第一个城市0（行号），它和其它城市的关系由矩阵第一行决定，而对于第二个城市1，它和其它城市的关系由第二行决定，以此类推。而且对于遍历过的城市无需再次遍历，即visited[i] = true。也就是当你遍历完第一行，你会发现第一列，包括对角线（\方向）都不用遍历了。
[1,1,0]
[1,1,0]
[0,0,1]
这里如果想通过把矩阵值置0的话，可以，不过需要多个置零（ij，ji，jj），且不是很容易理解，具体代码贴这了

class Solution {private int n;//行、列private boolean[] visited;//访问数组，长度为城市的个数，而不是二维矩阵public int findCircleNum(int[][] isConnected) {//初始化int count = 0;//计数n = isConnected.length;visited = new boolean[n];for(int i = 0; i < n; i++) {//每一行代表一个城市与其它城市的关系            if(visited[i]==false) {                dfs(isConnected, i); //每一轮dfs就刷掉一个省份，也就是把同个省的城市标记为truecount++;}            }return count;}    public void dfs(int[][] isConnected, int i) {visited[i] = true;//访问过的城市for(int j = 0; j < n; j++) {if(visited[j]==false && isConnected[i][j] == 1) dfs(isConnected, j);            }    }
}

130. 被围绕的区域

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思路一：DFS
被围绕的区域（为O的点）要求四周都是X（不包括对角线），这就要求该区域的点不能是边界点(第一行、第一列、最后一行、最后一列)；
正面思考就是判断哪些区域的点不包含边界点，然后填充该区域，但这样查找并不容易。如果反过来思考，查找哪些包含边界点的区域，就显得十分容易了，因为那些包含边界点的区域只需要通过遍历边界点就可以了。也就是遇到边界为O的点，就从该点出发，构造不可填充的区域（需要做标记），然后根据标记填充矩阵。

class Solution {private int m;private int n;//标记数组，标记不被填充的点，true的话代表不可以被填充private boolean[][] ox;public void solve(char[][] board) {m = board.length;n = board[0].length;ox = new boolean[m][n];//遍历边界点for(int i=0; i<m; i++) {for(int j=0; j<n ;j++) {//只需要遍历边界的点，从而找出那些不能被填充的点，然后做个标记就可以了。if(i==0 || j==0 || i==m-1 || j==n-1) {dfs(board, i, j);}}}//根据标记填充矩阵for(int i=0; i<m; i++) {for(int j=0; j<n ;j++) {                if(board[i][j]=='O' && !ox[i][j]) {board[i][j]='X';}}}}public void dfs(char[][] board, int x, int y) {if(x<0 || y<0 || x>=m || y>=n) return;//防止数组越界if(board[x][y] == 'X' || ox[x][y] == true)  return;//排除X和访问过的ox[x][y] = true;dfs(board, x-1, y);dfs(board, x+1, y);dfs(board, x, y-1);dfs(board, x, y+1);}
}

优化：如果不想使用标记数组，可对原数组进行修改，例如不需要填充设为‘A’，不过后面遍历时就需要将其改为’O’

417. 太平洋大西洋水流问题

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测试例子
[[1,1,1],[1,1,1],[1,1,1]]
[[1,2,3],[4,5,6],[7,8,9]]
超时例子
[[1,2,3,4,5,6,7,8,9,10,11,12,13,14],[52,53,54,55,56,57,58,59,60,61,62,63,64,15],[51,96,97,98,99,100,101,102,103,104,105,106,65,16],[50,95,132,133,134,135,136,137,138,139,140,107,66,17],[49,94,131,160,161,162,163,164,165,166,141,108,67,18],[48,93,130,159,180,181,182,183,184,167,142,109,68,19],[47,92,129,158,179,192,193,194,185,168,143,110,69,20],[46,91,128,157,178,191,196,195,186,169,144,111,70,21],[45,90,127,156,177,190,189,188,187,170,145,112,71,22],[44,89,126,155,176,175,174,173,172,171,146,113,72,23],[43,88,125,154,153,152,151,150,149,148,147,114,73,24],[42,87,124,123,122,121,120,119,118,117,116,115,74,25],[41,86,85,84,83,82,81,80,79,78,77,76,75,26],[40,39,38,37,36,35,34,33,32,31,30,29,28,27]]
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0,919,639,273,899,870,449,329,203,1005,14,275,514,507,671,519,745,725,443,792,692,963,293,865,128,899,819,93,658,21,151,140,930,132,756,171,992,195,491,537,542,847,155,46,344,816,907,431,883,340,565,565,646,849,772,116,738,581,199,738,592,692,868,864,814,614,25,148,151,858,675],[1684,696,172,71,382,978,968,156,203,651,711,758,287,902,872,393,630,443,934,358,25,616,569,231,420,173,247,558,667,95,575,340,133,738,754,858,58,712,356,604,705,57,352,982,950,567,717,922,1000,641,271,368,247,830,589,1009,345,178,557,1002,616,475,685,91,555,781,939,955,483,285,549,179,685,892,503,977,449,563,241,791,194,502,149,784,674,81,783,362,601,331,706,207,796,381,289,693,152,570,638,978],[1199,178,499,225,412,344,544,203,897,775,984,82,620,476,208,284,899,981,988,490,302,685,40,440,408,671,123,903,231,752,223,771,272,712,986,674,398,872,867,286,638,841,710,248,659,908,874,548,879,853,381,524,528,411,306,926,72,772,171,645,514,384,759,128,86,87,792,827,950,649,103,578,832,803,816,834,53,680,372,922,875,95,788,393,496,85,662,910,847,823,898,703,198,647,821,626,724,955,443,16],[1595,888,584,768,33,742,592,76,765,307,341,630,392,471,366,879,546,18,131,735,183,19,428,723,1008,591,340,723,536,125,81,472,356,656,231,379,740,813,798,495,110,129,116,493,590,824,714,479,832,835,204,357,845,975,71,195,556,753,908,435,220,980,897,566,978,118,936,708,274,724,546,726,195,1004,209,127,818,265,948,992,91,143,339,278,460,400,463,358,495,714,783,706,36,671,262,356,131,540,406,395],[1255,942,464,791,936,663,261,744,271,199,78,704,684,408,324,134,150,129,483,636,833,256,684,211,269,936,557,743,819,954,480,63,238,934,197,165,940,448,251,201,637,320,895,312,70,209,788,210,328,261,188,504,860,862,705,471,141,605,204,950,901,675,355,129,951,894,636,881,332,878,72,312,540,309,614,600,508,392,800,179,996,331,25,198,535,72,659,666,667,854,958,910,871,656,382,812,540,1008,36,215],[1229,98,517,758,398,473,348,248,207,491,769,193,164,784,381,689,847,383,698,856,579,646,757,792,644,129,594,175,479,620,380,698,61,239,446,801,702,137,391,899,618,151,435,772,277,158,803,466,883,491,313,452,480,412,234,114,883,171,631,352,133,1001,40,184,582,477,327,274,604,709,516,564,202,293,678,821,793,823,278,667,305,581,109,775,983,686,231,856,847,853,550,970,196,581,145,769,400,814,33,994],[1514,891,900,57,526,920,869,310,733,137,967,380,60,418,145,385,94,367,583,931,210,123,244,396,694,731,507,84,535,531,420,391,764,662,438,281,572,649,933,648,776,890,18,826,650,506,201,87,215,774,1008,767,888,242,153,924,315,1003,999,841,524,761,222,630,414,1002,253,328,642,176,966,760,408,327,577,49,823,120,126,28,237,476,785,467,61,280,733,366,273,722,549,139,826,761,112,230,754,355,548,738],[1523,857,488,920,174,407,311,339,518,427,357,97,246,484,906,649,754,629,1005,370,694,545,851,510,648,953,730,744,299,620,472,811,819,951,73,983,348,375,312,208,144,11,647,380,485,543,19,582,515,367,942,199,254,783,51,892,79,771,627,368,733,441,521,543,734,584,868,425,949,523,975,84,524,613,806,352,498,168,276,1003,525,560,544,769,685,585,1003,754,698,972,112,774,404,623,659,128,198,517,543,489],[1383,509,563,249,464,712,591,304,870,209,650,385,759,184,496,434,112,489,179,152,452,633,916,198,247,565,668,787,73,202,266,797,53,172,36,859,874,969,153,734,168,793,461,917,968,299,342,422,778,863,564,572,486,471,112,75,26,771,852,441,315,461,581,358,623,607,207,839,567,350,563,725,486,366,633,444,655,317,208,775,522,762,690,998,575,144,416,592,905,258,375,210,709,946,910,674,896,107,503,453],[1801,408,520,276,764,143,62,761,802,260,879,666,364,911,1007,930,45,413,864,941,661,581,493,713,870,394,377,756,491,871,199,633,269,709,251,376,195,303,479,339,905,348,996,260,249,993,180,285,396,34,568,399,605,51,102,465,435,470,563,269,683,752,244,942,804,486,660,341,779,130,22,675,468,360,277,60,343,447,687,81,823,245,823,770,638,915,578,416,727,131,675,752,226,261,37,20,737,687,703,859],[1808,715,524,618,418,143,20,103,932,697,175,745,284,340,505,912,597,73,670,315,547,335,409,763,587,436,125,666,114,818,515,263,875,381,223,283,514,233,729,436,920,246,523,546,576,19,449,163,434,109,820,971,787,220,724,716,646,839,724,750,999,230,356,217,601,569,842,458,793,561,884,55,797,398,592,363,759,383,869,535,834,679,497,963,889,563,669,878,745,736,970,86,956,316,293,899,876,478,347,11],[1382,574,56,169,314,638,874,415,11,85,940,836,755,427,789,986,333,801,854,68,879,167,496,825,473,780,66,691,600,756,692,971,320,91,482,976,719,347,381,73,774,311,251,519,81,382,848,756,525,44,814,394,201,300,561,665,422,618,698,364,364,381,326,26,814,798,992,523,487,363,938,252,16,179,113,439,904,951,185,419,986,341,804,529,984,707,184,748,315,873,103,21,596,419,37,400,559,371,265,389]]

思路一：暴力DFS–不推荐力扣133ms，思路二3ms

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