20190224

code/lc171.java

+package code;
+/*
+ * 171. Excel Sheet Column Number
+ * 题意：Excel列的表示转换为数字
+ * 难度：Easy
+ * 分类：Math
+ * 思路：注意AA代表的是27， 没有0的表示。26进位不是27。
+ * Tips：
+ */
+public class lc171 {
+    public int titleToNumber(String s) {
+        char[] ch_arr = s.toCharArray();
+        int res = 0;
+        for(int i=0; i<ch_arr.length; i++){
+            res += ( (ch_arr[i]-'A'+1) * Math.pow(26, ch_arr.length-1-i) );
+        }
+        return res;
+    }
+}

code/lc172.java

+package code;
+/*
+ * 172. Factorial Trailing Zeroes
+ * 题意：求n!末尾有几个0
+ * 难度：Easy
+ * 分类：Math
+ * 思路：思路记一下。如果末尾为0，则一定是2*5得到的0，2的数量一定比5多，所以只考虑5的数量，25算两个5。
+ * Tips：题目看似简单，没做过的话不好想
+ */
+public class lc172 {
+    public int trailingZeroes(int n) {
+        if(n==0) return 0;
+        return n/5 + trailingZeroes(n/5);
+    }
+}

code/lc189.java

+package code;
+/*
+ * 189. Rotate Array
+ * 题意：数组向后移几位，超出末尾的补到前边
+ * 难度：Easy
+ * 分类：Array
+ * 思路：一种换状替换，别忘了可能是多个环。
+ *      reverse 的方法，先整体反转，再按照k划分成两个数组分别反转
+ * Tips：
+ */
+public class lc189 {
+    public void rotate(int[] nums, int k) {
+        if(nums.length<2) return;
+        int sum = 0;
+        for (int start_index = 0; start_index < nums.length ; start_index++) {  //可能多个环，用sum判断是否停止
+            if(sum==nums.length) break;
+            int curr_index = start_index;
+            int next_index = (start_index+k)%nums.length;
+            int temp1 = nums[start_index];
+            int temp2 =0;
+            while(next_index!=start_index){
+                sum++;
+                temp2 = nums[next_index];
+                nums[next_index] = temp1;
+                temp1 =temp2;
+                curr_index = next_index;
+                next_index = (curr_index+k)%nums.length;
+            }
+            nums[start_index] = temp1;
+            sum++;
+        }
+    }
+
+    public void rotate2(int[] nums, int k) {
+        if(nums.length<2) return;
+        k = k%nums.length;      //处理k>length的case
+        reverse(nums, 0, nums.length-1);
+        reverse(nums, 0, k-1);
+        reverse(nums, k, nums.length-1);
+    }
+    public void reverse(int[] nums, int begin, int end){
+        while(begin<end){
+            int temp = nums[begin];
+            nums[begin] = nums[end];
+            nums[end] = temp;
+            begin++;
+            end--;
+        }
+        return;
+    }
+}

code/lc207.java

@@ -7,7 +7,7 @@ import java.util.*;
  * 题意：课程是否能够完成
  * 难度：Medium
  * 分类：Depth-first Search, Breadth-first Search, Graph, Topology Sort
- * 思路：两种方法，一种BFS拓扑排序，另一种DFS找是否有环
+ * 思路：两种方法，一种BFS拓扑排序(每个节点，先求出入度)，另一种DFS找是否有环
  * Tips：很经典的题，拓扑排序，判断图是否有环的DFS
  */
 public class lc207 {
@@ -17,8 +17,8 @@ public class lc207 {
         System.out.println(canFinish2(2, prerequisites));
     }
     public static boolean canFinish(int numCourses, int[][] prerequisites) {
-        int[] indegree = new int[numCourses];
-        int[][] graph = new int[numCourses][numCourses];
+        int[] indegree = new int[numCourses];   //计算入度
+        int[][] graph = new int[numCourses][numCourses];    //邻接矩阵
         for (int i = 0; i < prerequisites.length ; i++) {
             int node1 = prerequisites[i][0];
             int node2 = prerequisites[i][1];

code/lc210.java

+package code;
+
+import java.util.ArrayDeque;
+import java.util.Queue;
+/*
+ * 210. Course Schedule II
+ * 题意：课程是否能够完成
+ * 难度：Medium
+ * 分类：Depth-first Search, Breadth-first Search, Graph, Topology Sort
+ * 思路：两种方法，一种BFS拓扑排序(每个节点，先求出入度)，另一种DFS找是否有环
+ * Tips：和lc207一模一样，换了个输出
+ *       注意先统计入度并转化为邻接矩阵，之后就好操作了
+ */
+public class lc210 {
+    public static void main(String[] args) {
+        int[][] prerequisites = {{1,0},{2,0},{3,1},{3,2}};
+        //System.out.println(canFinish(2, prerequisites));
+        System.out.println(findOrder(4, prerequisites));
+    }
+    public static int[] findOrder(int numCourses, int[][] prerequisites) {
+        int[] degrees = new int[numCourses];
+        int[][] graph = new int[numCourses][numCourses];
+        for (int i = 0; i < prerequisites.length ; i++) {
+            int course1 = prerequisites[i][0];
+            int course2 = prerequisites[i][1];
+            graph[course2][course1] = 1;
+            degrees[course1]++;
+        }
+        Queue<Integer> qu = new ArrayDeque();
+        int[] res = new int[numCourses];
+        int sum = 0;
+        for (int i = 0; i < degrees.length ; i++) {
+            if(degrees[i]==0) {
+                qu.add(i);
+                res[sum] = i;
+                sum++;
+            }
+        }
+        while(!qu.isEmpty()){
+            int curr_course = qu.remove();
+            for (int i = 0; i < numCourses ; i++) {
+                if(graph[curr_course][i]==1){
+                    degrees[i]--;
+                    if(degrees[i]==0){
+                        qu.add(i);
+                        res[sum] = i;
+                        sum++;
+                    }
+                }
+            }
+        }
+        return sum==numCourses ? res : new int[0];
+    }
+}

readme.md

@@ -106,12 +106,16 @@ Language: Java
 | 152 [Java](./code/lc152.java)
 | 155 [Java](./code/lc155.java)
 | 160 [Java](./code/lc160.java)
+| 162 [Java](./code/lc162.java)
+| 166 [Java](./code/lc166.java)
 | 169 [Java](./code/lc169.java)
+| 179 [Java](./code/lc179.java)
 | 198 [Java](./code/lc198.java)
 | 200 [Java](./code/lc200.java)
 | 206 [Java](./code/lc206.java)
 | 207 [Java](./code/lc207.java)
 | 208 [Java](./code/lc208.java)
+| 212 [Java](./code/lc212.java)
 | 215 [Java](./code/lc215.java)
 | 221 [Java](./code/lc221.java)
 | 226 [Java](./code/lc226.java)