20190221

Author: mJackie

code/lc138.java

@@ -0,0 +1,43 @@
+package code;
+/*
+ * 138. Copy List with Random Pointer
+ * 题意：链表除了next指针外，还有一个random指针。拷贝这个链表
+ * 难度：Medium
+ * 分类：Hash Table, LinkedList
+ * 思路：一种用HashTable，把源节点和复制的节点对应起来，空间复杂度为O(n)
+ *      把要复制的节点直接连接在该节点之后，操作完以后再分开
+ * Tips：和139做对比，因为需要每步骤返回值，所以递归的方法在这题更合适
+ */
+public class lc138 {
+    class RandomListNode {
+        int label;
+        RandomListNode next, random;
+        RandomListNode(int x) { this.label = x; }
+    };
+    public RandomListNode copyRandomList(RandomListNode head) {
+        if(head==null) return null;
+        RandomListNode node = head;
+        while(node!=null){
+            RandomListNode temp = new RandomListNode(node.label);
+            temp.next = node.next;
+            node.next = temp;
+            node = temp.next;
+        }
+        node = head;
+        while(node!=null){
+            if(node.random!=null) node.next.random = node.random.next;  //注意判断是否为空
+            node = node.next.next;
+        }
+        node = head;
+        RandomListNode res = head.next;
+        RandomListNode temp = head.next;
+        while(temp.next!=null){
+            node.next = temp.next;
+            temp.next = node.next.next;
+            node = node.next;
+            temp = temp.next;
+        }
+        node.next = null;   //别忘了最后一个节点连接给null
+        return res;
+    }
+}

code/lc149.java

@@ -0,0 +1,53 @@
+package code;
+
+import java.util.HashMap;
+
+/*
+ * 149. Max Points on a Line
+ * 题意：给定多个点，问一条线上最多几个点
+ * 难度：Hard
+ * 分类：HashTable, Math
+ * 思路：每次选出一个点，与其他点匹配。因为斜率是小数，所以用两个整数（双层HashMap）来表示小数。时间O(n^2)
+ * Tips：
+ */
+public class lc149 {
+    class Point {
+        int x;
+        int y;
+        Point() { x = 0; y = 0; }
+        Point(int a, int b) { x = a; y = b; }
+    }
+    public int maxPoints(Point[] points) {
+        if(points.length<=2) return points.length;
+        int res = 0;
+        for (int i = 0; i < points.length-1 ; i++) {
+            HashMap<Integer, HashMap<Integer, Integer>> hm = new HashMap<>();
+            int overlap = 1;
+            int max = 0;
+            for (int j = i+1; j < points.length ; j++) {
+                int x = points[i].x - points[j].x;
+                int y = points[i].y - points[j].y;
+                if(x==0&&y==0){ //重复点
+                    overlap++;
+                    continue;
+                }
+                int gcd = genGCD(x, y);
+                if(gcd!=0){ //注意，当一个值为0时，gcd==0
+                    x = x/gcd;
+                    y = y/gcd;
+                }
+                HashMap<Integer, Integer> hm2 = hm.getOrDefault(x, new HashMap<Integer, Integer>());
+                hm2.put(y,hm2.getOrDefault(y,0)+1);
+                hm.put(x,hm2);
+                max = Math.max(max, hm2.get(y));
+            }
+            res = Math.max(res, max+overlap);   //里边循环完了，才知道有几个overlap，所以在循环外加
+        }
+        return res;
+    }
+
+    public int genGCD(int a, int b){
+        if(b==0) return a;  //注意和GCD不一样的是这没有比较大小，交换a,b。因为可能有负数
+        return genGCD(b, a%b);
+    }
+}

code/lc150.java

@@ -0,0 +1,41 @@
+package code;
+/*
+ * 150. Evaluate Reverse Polish Notation
+ * 题意：波兰表达式运算
+ * 难度：Medium
+ * 分类：Stack
+ * 思路：用栈即可
+ * Tips：
+ */
+import java.util.Stack;
+
+public class lc150 {
+    public int evalRPN(String[] tokens) {
+        if(tokens.length==1) return Integer.valueOf(tokens[0]);
+        Stack<Integer> st = new Stack();
+        String opration = "+-*/";
+        int res = 0;
+        for (int i = 0; i < tokens.length ; i++) {
+            if(opration.contains(tokens[i])){
+                int n2 = st.pop();  //注意n1,n2顺序，除法时两个数是有顺序的
+                int n1 = st.pop();
+                System.out.println(n1);
+                System.out.println(n2);
+                if(tokens[i].equals("+")){  //注意用equeals
+                    res = n1 + n2;
+                }else if(tokens[i].equals("-")){
+                    res = n1 - n2;
+                }else if(tokens[i].equals("*")){
+                    res = n1 * n2;
+                }else if(tokens[i].equals("/")){
+                    res = n1 / n2;
+                }
+                st.push(res);
+                System.out.println(res);
+            }else{
+                st.push(Integer.valueOf(tokens[i]));;
+            }
+        }
+        return res;
+    }
+}

readme.md

@@ -92,13 +92,17 @@ Language: Java
 | 131 [Java](./code/lc131.java)
 | 134 [Java](./code/lc134.java)
 | 136 [Java](./code/lc136.java)
+| 138 [Java](./code/lc138.java)
 | 139 [Java](./code/lc139.java)
+| 140 [Java](./code/lc140.java)
 | 141 [Java](./code/lc141.java)
 | 142 [Java](./code/lc142.java)
 | 144 [Java](./code/lc144.java)
 | 145 [Java](./code/lc145.java)
 | 146 [Java](./code/lc146.java)
 | 148 [Java](./code/lc148.java)
+| 149 [Java](./code/lc149.java)
+| 150 [Java](./code/lc150.java)
 | 152 [Java](./code/lc152.java)
 | 155 [Java](./code/lc155.java)
 | 160 [Java](./code/lc160.java)