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mJackiemJackie
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20190224
1 parent 3b59391 commit 94cfc5b

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6 files changed

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-3
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6 files changed

+146
-3
lines changed

Diff for: ‎code/lc171.java

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

Diff for: ‎code/lc172.java

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package code;
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/*
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* 172. Factorial Trailing Zeroes
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* 题意：求n!末尾有几个0
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* 难度：Easy
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* 分类：Math
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* 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。
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* Tips：题目看似简单，没做过的话不好想
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*/
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public class lc172 {
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public int trailingZeroes(int n) {
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if(n==0) return 0;
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return n/5 + trailingZeroes(n/5);
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}
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}

Diff for: ‎code/lc189.java

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

Diff for: ‎code/lc207.java

+3-3
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@@ -7,7 +7,7 @@
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* 题意：课程是否能够完成
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* 难度：Medium
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* 分类：Depth-first Search, Breadth-first Search, Graph, Topology Sort
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* 思路：两种方法，一种BFS拓扑排序，另一种DFS找是否有环
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* 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
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* Tips：很经典的题，拓扑排序，判断图是否有环的DFS
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*/
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public class lc207 {
@@ -17,8 +17,8 @@ public static void main(String[] args) {
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System.out.println(canFinish2(2, prerequisites));
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}
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public static boolean canFinish(int numCourses, int[][] prerequisites) {
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int[] indegree = new int[numCourses];
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int[][] graph = new int[numCourses][numCourses];
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int[] indegree = new int[numCourses]; //计算入度
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int[][] graph = new int[numCourses][numCourses]; //邻接矩阵
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for (int i = 0; i < prerequisites.length ; i++) {
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int node1 = prerequisites[i][0];
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int node2 = prerequisites[i][1];

Diff for: ‎code/lc210.java

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

Diff for: ‎readme.md

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

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