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CyC2018
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/95903878-725b-4ed9-bded-bc4aae0792a9.jpg"/> </div><br>
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广度优先搜索一层一层地进行遍历，每层遍历都以上一层遍历的结果作为起点，遍历一个距离能访问到的所有节点。需要注意的是，遍历过的节点不能再次被遍历。
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广度优先搜索一层一层地进行遍历，每层遍历都是以上一层遍历的结果作为起点，遍历一个距离能访问到的所有节点。需要注意的是，遍历过的节点不能再次被遍历。
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第一层：
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@ -75,33 +75,39 @@
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题目描述：0 表示可以经过某个位置，求解从左上角到右下角的最短路径长度。
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```java
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public int shortestPathBinaryMatrix(int[][] grids) {
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int[][] direction = {{1, -1}, {1, 0}, {1, 1}, {0, -1}, {0, 1}, {-1, -1}, {-1, 0}, {-1, 1}};
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int m = grids.length, n = grids[0].length;
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Queue<Pair<Integer, Integer>> queue = new LinkedList<>();
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queue.add(new Pair<>(0, 0));
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int pathLength = 0;
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while (!queue.isEmpty()) {
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int size = queue.size();
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pathLength++;
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while (size-- > 0) {
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Pair<Integer, Integer> cur = queue.poll();
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int cr = cur.getKey(), cc = cur.getValue();
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grids[cr][cc] = 1; // 标记
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for (int[] d : direction) {
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int nr = cr + d[0], nc = cc + d[1];
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if (nr < 0 || nr >= m || nc < 0 || nc >= n || grids[nr][nc] == 1) {
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public int shortestPathBinaryMatrix(int[][] grids) {
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if (grids == null || grids.length == 0 || grids[0].length == 0) {
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return -1;
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}
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int[][] direction = {{1, -1}, {1, 0}, {1, 1}, {0, -1}, {0, 1}, {-1, -1}, {-1, 0}, {-1, 1}};
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int m = grids.length, n = grids[0].length;
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Queue<Pair<Integer, Integer>> queue = new LinkedList<>();
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queue.add(new Pair<>(0, 0));
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int pathLength = 0;
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while (!queue.isEmpty()) {
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int size = queue.size();
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pathLength++;
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while (size-- > 0) {
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Pair<Integer, Integer> cur = queue.poll();
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int cr = cur.getKey(), cc = cur.getValue();
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if (grids[cr][cc] == 1) {
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continue;
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}
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if (nr == m - 1 && nc == n - 1) {
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return pathLength + 1;
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if (cr == m - 1 && cc == n - 1) {
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return pathLength;
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}
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grids[cr][cc] = 1; // 标记
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for (int[] d : direction) {
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int nr = cr + d[0], nc = cc + d[1];
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if (nr < 0 || nr >= m || nc < 0 || nc >= n) {
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continue;
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}
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queue.add(new Pair<>(nr, nc));
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}
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queue.add(new Pair<>(nr, nc));
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}
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}
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return -1;
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}
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return -1;
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}
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```
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## 2. 组成整数的最小平方数数量
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@ -34,7 +34,7 @@
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/95903878-725b-4ed9-bded-bc4aae0792a9.jpg"/> </div><br>
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广度优先搜索一层一层地进行遍历，每层遍历都以上一层遍历的结果作为起点，遍历一个距离能访问到的所有节点。需要注意的是，遍历过的节点不能再次被遍历。
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广度优先搜索一层一层地进行遍历，每层遍历都是以上一层遍历的结果作为起点，遍历一个距离能访问到的所有节点。需要注意的是，遍历过的节点不能再次被遍历。
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第一层：
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@ -75,33 +75,39 @@
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题目描述：0 表示可以经过某个位置，求解从左上角到右下角的最短路径长度。
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```java
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public int shortestPathBinaryMatrix(int[][] grids) {
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int[][] direction = {{1, -1}, {1, 0}, {1, 1}, {0, -1}, {0, 1}, {-1, -1}, {-1, 0}, {-1, 1}};
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int m = grids.length, n = grids[0].length;
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Queue<Pair<Integer, Integer>> queue = new LinkedList<>();
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queue.add(new Pair<>(0, 0));
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int pathLength = 0;
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while (!queue.isEmpty()) {
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int size = queue.size();
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pathLength++;
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while (size-- > 0) {
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Pair<Integer, Integer> cur = queue.poll();
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int cr = cur.getKey(), cc = cur.getValue();
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grids[cr][cc] = 1; // 标记
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for (int[] d : direction) {
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int nr = cr + d[0], nc = cc + d[1];
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if (nr < 0 || nr >= m || nc < 0 || nc >= n || grids[nr][nc] == 1) {
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public int shortestPathBinaryMatrix(int[][] grids) {
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if (grids == null || grids.length == 0 || grids[0].length == 0) {
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return -1;
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}
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int[][] direction = {{1, -1}, {1, 0}, {1, 1}, {0, -1}, {0, 1}, {-1, -1}, {-1, 0}, {-1, 1}};
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int m = grids.length, n = grids[0].length;
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Queue<Pair<Integer, Integer>> queue = new LinkedList<>();
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queue.add(new Pair<>(0, 0));
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int pathLength = 0;
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while (!queue.isEmpty()) {
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int size = queue.size();
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pathLength++;
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while (size-- > 0) {
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Pair<Integer, Integer> cur = queue.poll();
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int cr = cur.getKey(), cc = cur.getValue();
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if (grids[cr][cc] == 1) {
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continue;
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}
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if (nr == m - 1 && nc == n - 1) {
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return pathLength + 1;
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if (cr == m - 1 && cc == n - 1) {
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return pathLength;
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}
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grids[cr][cc] = 1; // 标记
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for (int[] d : direction) {
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int nr = cr + d[0], nc = cc + d[1];
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if (nr < 0 || nr >= m || nc < 0 || nc >= n) {
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continue;
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}
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queue.add(new Pair<>(nr, nc));
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}
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queue.add(new Pair<>(nr, nc));
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}
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}
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return -1;
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}
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return -1;
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}
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```
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## 2. 组成整数的最小平方数数量
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