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CyC2018
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* [BST](#bst)
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* [Trie](#trie)
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* [图](#图)
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* [拓扑排序](#拓扑排序)
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* [并查集](#并查集)
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* [位运算](#位运算)
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* [参考资料](#参考资料)
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<!-- GFM-TOC -->
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@ -3447,14 +3449,14 @@ public int majorityElement(int[] nums) {
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}
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```
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可以利用 Boyer-Moore Majority Vote Algorithm 来解决这个问题，使得时间复杂度为 O(N)。可以这么理解该算法：使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素不相等时，令 cnt--。如果前面查找了 i 个元素，且 cnt == 0 ，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
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可以利用 Boyer-Moore Majority Vote Algorithm 来解决这个问题，使得时间复杂度为 O(N)。可以这么理解该算法：使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素不相等时，令 cnt--。如果前面查找了 i 个元素，且 cnt == 0，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2，因为如果多于 i / 2 的话 cnt 就一定不会为 0。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
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```java
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public int majorityElement(int[] nums) {
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int cnt = 1, majority = nums[0];
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for (int i = 1; i < nums.length; i++) {
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majority = (cnt == 0) ? nums[i] : majority;
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cnt = (majority == nums[i]) ? cnt + 1 : cnt - 1;
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int cnt = 0, majority = nums[0];
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for (int num : nums) {
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majority = (cnt == 0) ? num : majority;
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cnt = (majority == num) ? cnt + 1 : cnt - 1;
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}
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return majority;
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}
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@ -5912,6 +5914,122 @@ class MapSum {
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## 图
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### 拓扑排序
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常用于在具有先序关系的任务规划中。
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**课程安排的合法性**
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[Leetcode : 207. Course Schedule (Medium)](https://leetcode.com/problems/course-schedule/description/)
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```html
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2, [[1,0]]
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return true
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```
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```html
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2, [[1,0],[0,1]]
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return false
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```
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题目描述：一个课程可能会先修课程，判断给定的先修课程规定是否合法。
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本题不需要使用拓扑排序，只需要检测有向图是否存在环即可。
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```java
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public boolean canFinish(int numCourses, int[][] prerequisites) {
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List<Integer>[] graphic = new List[numCourses];
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for (int i = 0; i < numCourses; i++)
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graphic[i] = new ArrayList<>();
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for (int[] pre : prerequisites)
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graphic[pre[0]].add(pre[1]);
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boolean[] globalMarked = new boolean[numCourses];
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boolean[] localMarked = new boolean[numCourses];
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for (int i = 0; i < numCourses; i++)
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if (!dfs(globalMarked, localMarked, graphic, i))
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return false;
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return true;
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}
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private boolean dfs(boolean[] globalMarked, boolean[] localMarked, List<Integer>[] graphic, int curNode) {
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if (localMarked[curNode])
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return false;
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if (globalMarked[curNode])
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return true;
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globalMarked[curNode] = true;
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localMarked[curNode] = true;
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for (int nextNode : graphic[curNode])
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if (!dfs(globalMarked, localMarked, graphic, nextNode))
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return false;
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localMarked[curNode] = false;
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return true;
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}
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```
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**课程安排的顺序**
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[Leetcode : 210. Course Schedule II (Medium)](https://leetcode.com/problems/course-schedule-ii/description/)
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```html
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4, [[1,0],[2,0],[3,1],[3,2]]
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There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].
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```
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使用 DFS 来实现拓扑排序，使用一个栈存储后序遍历结果，这个栈元素的逆序结果就是拓扑排序结果。
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证明：对于任何先序关系：v->w，后序遍历结果可以保证 w 先进入栈中，因此栈的逆序结果中 v 会在 w 之前。
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```java
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public int[] findOrder(int numCourses, int[][] prerequisites) {
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List<Integer>[] graphic = new List[numCourses];
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for (int i = 0; i < numCourses; i++)
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graphic[i] = new ArrayList<>();
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for (int[] pre : prerequisites)
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graphic[pre[0]].add(pre[1]);
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Stack<Integer> topologyOrder = new Stack<>();
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boolean[] globalMarked = new boolean[numCourses];
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boolean[] localMarked = new boolean[numCourses];
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for (int i = 0; i < numCourses; i++)
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if (!dfs(globalMarked, localMarked, graphic, i, topologyOrder))
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return new int[0];
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int[] ret = new int[numCourses];
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for (int i = numCourses - 1; i >= 0; i--)
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ret[i] = topologyOrder.pop();
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return ret;
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}
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private boolean dfs(boolean[] globalMarked, boolean[] localMarked, List<Integer>[] graphic, int curNode, Stack<Integer> topologyOrder) {
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if (localMarked[curNode])
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return false;
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if (globalMarked[curNode])
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return true;
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globalMarked[curNode] = true;
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localMarked[curNode] = true;
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for (int nextNode : graphic[curNode])
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if (!dfs(globalMarked, localMarked, graphic, nextNode, topologyOrder))
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return false;
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localMarked[curNode] = false;
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topologyOrder.push(curNode);
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return true;
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}
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```
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### 并查集
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并查集可以动态地连通两个点，并且可以非常快速地判断两个点是否连通。
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**冗余连接**
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[Leetcode : 684. Redundant Connection (Medium)](https://leetcode.com/problems/redundant-connection/description/)
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