20190227

mJackie · mJackie · commit 6e70e3a7f729 · 2019-02-27T10:26:35.000+08:00
diff --git a/code/lc218.java b/code/lc218.java
@@ -0,0 +1,53 @@
+package code;
+
+import java.util.*;
+/*
+ * 218. The Skyline Problem
+ * 题意：高楼轮廓
+ * 难度：Hard
+ * 分类：Divide and Conquer, Heap, Binary indexed Tree, Segment Tree
+ * 思路：转化为坐标，按x排序。左端点add，右端点remove，用优先队列输出y坐标，如果y更改了就输出。
+ * Tips：
+ */
+public class lc218 {
+    public static void main(String[] args) {
+        int[][] buildings = {{0,2,3}, {2,5,3}};
+        getSkyline(buildings);
+    }
+    public static List<int[]> getSkyline(int[][] buildings) {
+        List<int[]> res = new ArrayList<>();
+        int[][] arr = new int[buildings.length*2][2];
+        for (int i = 0, j=0; i < buildings.length ; i++) {  //转换成坐标
+            arr[j][0] = buildings[i][0];
+            arr[j][1] = buildings[i][2];
+            j++;
+            arr[j][0] = buildings[i][1];
+            arr[j][1] = -buildings[i][2];   //结束节点y用负数表示
+            j++;
+        }
+        Arrays.sort(arr, new Comparator<int[]>() {  //定义比较方法
+            @Override
+            public int compare(int[] o1, int[] o2) {
+                if(o1[0]!=o2[0])
+                    return o1[0] - o2[0];
+                else
+                    return o2[1] - o1[1];       //注意这一点，防止 {{0,2,3}, {2,5,3}} case。 两个点重合了不会连续输出。
+            }
+        });
+        PriorityQueue<Integer> pr = new PriorityQueue<Integer>();
+        pr.add(0);
+        int pre = 0;
+        for (int i = 0; i < arr.length ; i++) {
+            if(arr[i][1]>0){
+                pr.add(-arr[i][1]);
+            }else{
+                pr.remove(arr[i][1]);
+            }
+            if(pr.peek()!=pre){
+                res.add(new int[]{arr[i][0], -pr.peek()});
+                pre = pr.peek();
+            }
+        }
+        return res;
+    }
+}
diff --git a/code/lc227.java b/code/lc227.java
@@ -0,0 +1,48 @@
+package code;
+
+import java.util.Stack;
+
+/*
+ * 227. Basic Calculator II
+ * 题意：表达式计算
+ * 难度：Medium
+ * 分类：String
+ * 思路：很巧妙的方法，每次遍历到下一个符号的时候，计算前一个符号的运算，泥面膜了复杂的逻辑。
+ *      + - 运算直接入栈，* / 运算则计算后将结果入栈，实现了 * / 优先运算
+ * Tips：自己想的方法会非常麻烦。这个解法非常聪明。
+ */
+public class lc227 {
+    public int calculate(String s) {
+        char[] chs = s.replace(" ","").toCharArray();
+        int num = 0;
+        char sign = '+';
+        Stack<Integer> st = new Stack();
+        for (int i = 0; i < chs.length ; i++) {
+            if(Character.isDigit(chs[i])){
+                num = num * 10 + chs[i]-'0';
+            }
+            if( !Character.isDigit(chs[i]) || i==chs.length-1 ){    //便利到最后，即使不是符号，也要计算
+                if(sign=='+'){
+                    st.push(num);
+                }
+                else if(sign=='-'){
+                    st.push(-num);
+                }
+                else if(sign=='*'){
+                    st.push(st.pop()*num);
+                }
+                else if(sign=='/'){
+                    st.push(st.pop()/num);
+                }
+                num = 0;
+                sign = chs[i];  //非常聪明
+            }
+        }
+        int res = 0;
+        for(Integer i : st){
+            System.out.println(i);
+            res += i;
+        }
+        return res;
+    }
+}
diff --git a/code/lc230.java b/code/lc230.java
@@ -0,0 +1,37 @@
+package code;
+
+import java.util.Stack;
+
+/*
+ * 230. Kth Smallest Element in a BST
+ * 题意：二叉搜索树种第k小的数
+ * 难度：Medium
+ * 分类：Binary Search, Tree
+ * 思路：每次找到最小的值，k--。中序遍历。
+ * Tips：非递归中序遍历还是不熟练。。。
+ */
+public class lc230 {
+    public class TreeNode {
+        int val;
+        TreeNode left;
+        TreeNode right;
+
+        TreeNode(int x) {
+            val = x;
+        }
+    }
+    public int kthSmallest(TreeNode root, int k) {
+        Stack<TreeNode> st = new Stack();
+        while(!st.isEmpty()||root!=null){
+            while(root!=null) {
+                st.add(root);
+                root = root.left;
+            }
+            root = st.pop();
+            k--;
+            if(k==0) return root.val;
+            root = root.right;
+        }
+        return 0;
+    }
+}
diff --git a/code/lc289.java b/code/lc289.java
@@ -0,0 +1,49 @@
+package code;
+/*
+ * 289. Game of Life
+ * 题意：按照游戏规则，计算下一个时刻的矩阵，inpalce
+ * 难度：Medium
+ * 分类：Array
+ * 思路：和lc130的trick很类似，先替换为其他值，最后再置回
+ *      需要注意几点，用2-bit表示状态转移，用&和位移操作省时
+ *      需要更新下规则与现在的表示一致
+ * Tips：
+ */
+public class lc289 {
+    public void gameOfLife(int[][] board) {
+        if (board == null || board.length == 0) return;
+        int m = board.length, n = board[0].length;
+
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                int lives = liveNeighbors(board, m, n, i, j);
+
+                // In the beginning, every 2nd bit is 0;
+                // So we only need to care about when will the 2nd bit become 1.
+                if (board[i][j] == 1 && lives >= 2 && lives <= 3) {
+                    board[i][j] = 3; // Make the 2nd bit 1: 01 ---> 11
+                }
+                if (board[i][j] == 0 && lives == 3) {
+                    board[i][j] = 2; // Make the 2nd bit 1: 00 ---> 10
+                }
+            }
+        }
+
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                board[i][j] >>= 1;  // Get the 2nd state.
+            }
+        }
+    }
+
+    public int liveNeighbors(int[][] board, int m, int n, int i, int j) {
+        int lives = 0;
+        for (int x = Math.max(i - 1, 0); x <= Math.min(i + 1, m - 1); x++) {    // 用max,min判断边界
+            for (int y = Math.max(j - 1, 0); y <= Math.min(j + 1, n - 1); y++) {
+                lives += board[x][y] & 1;
+            }
+        }
+        lives -= board[i][j] & 1;   //减去自己
+        return lives;
+    }
+}
diff --git a/code/lc295.java b/code/lc295.java
@@ -0,0 +1,28 @@
+package code;
+
+import java.util.PriorityQueue;
+
+public class lc295 {
+    class MedianFinder {
+        PriorityQueue<Integer> pq1;    //默认是最小，右半边
+        PriorityQueue<Integer> pq2;    //左半边
+
+        /** initialize your data structure here. */
+        public MedianFinder() {
+            this.pq1 = new PriorityQueue();
+            this.pq2 = new PriorityQueue();
+        }
+
+        public void addNum(int num) {
+            pq1.add(num);   //两个队列都过一遍
+            pq2.add(-pq1.poll());
+            if (pq1.size() < pq2.size())    //如果中位数是一个数，就存在左半边
+                pq1.add(-pq2.poll());
+        }
+
+        public double findMedian() {
+            if(pq1.size()==pq2.size()+1) return pq1.peek();
+            return -((double)(-pq1.peek()+pq2.peek()))/2;
+        }
+    }
+}
diff --git a/code/lc315.java b/code/lc315.java
@@ -0,0 +1,62 @@
+package code;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+/*
+ * 315. Count of Smaller Numbers After Self
+ * 题意：给一个数组，计算这个数右边比这个数小的数的个数
+ * 难度：Hard
+ * 分类：Divide and Conquer, Binary indexed Tree, Segment Tree, Binary Search Tree
+ * 思路：两种思路，一种用二叉搜索树这类数据结构 https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76580/9ms-short-Java-BST-solution-get-answer-when-building-BST
+ *      一种归并排序的思路，归并的时候统计左右交换数目。如果一个数从这个数的右边交换到左边，则+1。因为有重复数字，所以用将ndex进行排序
+ *      https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76583/11ms-JAVA-solution-using-merge-sort-with-explanation
+ *      再有一种复杂度稍微高点的思路，从后往前插入排序，插入的时候二分搜索
+ *      https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/76576/My-simple-AC-Java-Binary-Search-code
+ * Tips：好难呀，我日！
+ */
+public class lc315 {
+    class TreeNode{
+        int val;
+        int dup_num;
+        int sum;
+        TreeNode left;
+        TreeNode right;
+        TreeNode(int val, int dup_num, int sum){
+            this.val = val;
+            this.dup_num = dup_num; //相同点的数目
+            this.sum = sum; //该节点左下节点个数，也就是比该节点值小的
+        }
+    }
+
+    TreeNode root;
+    public List<Integer> countSmaller(int[] nums) {
+        if(nums.length<1) return new ArrayList<>();
+        Integer[] res_arr = new Integer[nums.length];   //用res_arr保存结果，否则结束了还要遍历数来找结果
+        root = new TreeNode(nums[nums.length-1], 1, 0);
+        res_arr[nums.length-1] = 0;
+        for (int i = nums.length-2; i >=0 ; i--) {
+            insert(root, nums[i], res_arr, i, 0);
+        }
+        return Arrays.asList(res_arr);  //数组转换为list
+    }
+
+    public TreeNode insert(TreeNode tn, int n, Integer[] res_arr, int i, int path){ //path记录了路径上比该点小的节点的个数
+        if(tn==null) {
+            tn = new TreeNode(n, 1, 0);
+            res_arr[i] = path;
+            System.out.print(i);
+            System.out.println("----"+path);
+        }
+        else if(tn.val==n){
+            tn.dup_num++;
+            res_arr[i] = path + tn.sum;
+        }else if(tn.val>n){
+            tn.sum++;
+            tn.left = insert(tn.left, n, res_arr, i, path);
+        }else{
+            tn.right = insert(tn.right, n, res_arr, i, path + tn.dup_num + tn.sum);
+        }
+        return tn;  //递归，返回的结果为节点，供上层节点赋值
+    }
+}
diff --git a/readme.md b/readme.md
@@ -135,6 +135,7 @@ Language: Java
 | 301 [Java](./code/lc301.java)
 | 309 [Java](./code/lc309.java)
 | 312 [Java](./code/lc312.java)
+
 | 322 [Java](./code/lc322.java)
 | 337 [Java](./code/lc337.java)
 | 338 [Java](./code/lc338.java)
