4道题目

Author: luzhipeng

Diff for: 328.odd-even-linked-list.md

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+
+## 题目地址
+https://leetcode.com/problems/odd-even-linked-list/description/
+
+## 题目描述
+
+```
+Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.
+
+You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.
+
+Example 1:
+
+Input: 1->2->3->4->5->NULL
+Output: 1->3->5->2->4->NULL
+Example 2:
+
+Input: 2->1->3->5->6->4->7->NULL
+Output: 2->3->6->7->1->5->4->NULL
+Note:
+
+The relative order inside both the even and odd groups should remain as it was in the input.
+The first node is considered odd, the second node even and so on ...
+```
+
+## 思路
+符合直觉的想法是，先遍历一遍找出奇数的节点。然后再遍历一遍找出偶数节点，最后串起来。
+
+但是有两个问题，如果不修改节点的话，需要借助额外的空间，空间复杂度是N。如果修改的话，会对第二次遍历（遍历偶数节点）造成影响。
+
+因此可以采用一种做法：  遍历一次，每一步同时修改两个节点就好了，这样就可以规避上面两个问题。
+## 关键点解析
+
+- 用虚拟节点来简化操作
+
+- 循环的结束条件设置为 `odd && odd.next && even && even.next`, 不应该是`odd && even`, 否则需要记录一下奇数节点的最后一个节点，复杂了操作
+
+## 代码
+```js
+/*
+ * @lc app=leetcode id=328 lang=javascript
+ *
+ * [328] Odd Even Linked List
+ *
+ * https://leetcode.com/problems/odd-even-linked-list/description/
+ *
+ * algorithms
+ * Medium (48.22%)
+ * Total Accepted:    137.6K
+ * Total Submissions: 284.2K
+ * Testcase Example:  '[1,2,3,4,5]'
+ *
+ * Given a singly linked list, group all odd nodes together followed by the
+ * even nodes. Please note here we are talking about the node number and not
+ * the value in the nodes.
+ * 
+ * You should try to do it in place. The program should run in O(1) space
+ * complexity and O(nodes) time complexity.
+ * 
+ * Example 1:
+ * 
+ * 
+ * Input: 1->2->3->4->5->NULL
+ * Output: 1->3->5->2->4->NULL
+ * 
+ * 
+ * Example 2:
+ * 
+ * 
+ * Input: 2->1->3->5->6->4->7->NULL
+ * Output: 2->3->6->7->1->5->4->NULL
+ * 
+ * 
+ * Note:
+ * 
+ * 
+ * The relative order inside both the even and odd groups should remain as it
+ * was in the input.
+ * The first node is considered odd, the second node even and so on ...
+ * 
+ * 
+ */
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @return {ListNode}
+ */
+var oddEvenList = function(head) {
+    if (!head || !head.next) return head;
+
+    const dummyHead1 = {
+        next: head
+    }
+    const dummyHead2 = {
+        next: head.next
+    }
+
+    let odd = dummyHead1.next;
+    let even = dummyHead2.next;
+    
+    while(odd && odd.next && even && even.next) {
+        const oddNext = odd.next.next;
+        const evenNext = even.next.next;
+        
+        odd.next = oddNext;
+        even.next = evenNext;
+
+        odd = oddNext;
+        even = evenNext;
+    }
+
+    odd.next = dummyHead2.next;
+
+    return dummyHead1.next;
+
+};
+```
+

Diff for: 349.intersection-of-two-arrays.md

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+
+## 题目地址
+https://leetcode.com/problems/intersection-of-two-arrays/description/
+
+## 题目描述
+
+```
+Given two arrays, write a function to compute their intersection.
+
+Example 1:
+
+Input: nums1 = [1,2,2,1], nums2 = [2,2]
+Output: [2]
+Example 2:
+
+Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
+Output: [9,4]
+Note:
+
+Each element in the result must be unique.
+The result can be in any order.
+
+```
+
+## 思路
+
+先遍历第一个数组，将其存到hashtable中，
+然后遍历第二个数组，如果在hashtable中存在就push到return，然后清空hashtable即可。
+
+## 关键点解析
+
+无
+
+## 代码
+```js
+/*
+ * @lc app=leetcode id=349 lang=javascript
+ *
+ * [349] Intersection of Two Arrays
+ *
+ * https://leetcode.com/problems/intersection-of-two-arrays/description/
+ *
+ * algorithms
+ * Easy (53.11%)
+ * Total Accepted:    203.6K
+ * Total Submissions: 380.9K
+ * Testcase Example:  '[1,2,2,1]\n[2,2]'
+ *
+ * Given two arrays, write a function to compute their intersection.
+ * 
+ * Example 1:
+ * 
+ * 
+ * Input: nums1 = [1,2,2,1], nums2 = [2,2]
+ * Output: [2]
+ * 
+ * 
+ * 
+ * Example 2:
+ * 
+ * 
+ * Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
+ * Output: [9,4]
+ * 
+ * 
+ * Note:
+ * 
+ * 
+ * Each element in the result must be unique.
+ * The result can be in any order.
+ * 
+ * 
+ * 
+ * 
+ */
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number[]}
+ */
+var intersection = function(nums1, nums2) {
+    const visited = {};
+    const ret = [];
+    for(let i = 0; i < nums1.length; i++) {
+        const num = nums1[i];
+
+        visited[num] = num;
+    }
+
+    for(let i = 0; i < nums2.length; i++) {
+        const num = nums2[i];
+
+        if (visited[num] !== undefined) {
+            ret.push(num);
+            visited[num] = undefined;
+        }
+    }
+
+    return ret;
+
+};
+```
+

Diff for: 445.add-two-numbers-ii.md

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+
+## 题目地址
+https://leetcode.com/problems/add-two-numbers-ii/description/
+
+## 题目描述
+
+```
+You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
+
+You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+
+Follow up:
+What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
+
+Example:
+
+Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
+Output: 7 -> 8 -> 0 -> 7
+
+```
+## 思路
+
+由于需要从低位开始加，然后进位。 因此可以采用栈来简化操作。
+依次将两个链表的值分别入栈stack1和stack2，然后相加入栈stack，进位操作用一个变量carried记录即可。
+
+最后根据stack生成最终的链表即可。
+
+## 关键点解析
+
+- 栈的基本操作
+- carried 变量记录进位
+- 循环的终止条件设置成`stack.length > 0` 可以简化操作
+- 注意特殊情况， 比如 1 + 99 = 100
+
+## 代码
+```js
+/*
+ * @lc app=leetcode id=445 lang=javascript
+ *
+ * [445] Add Two Numbers II
+ *
+ * https://leetcode.com/problems/add-two-numbers-ii/description/
+ *
+ * algorithms
+ * Medium (49.31%)
+ * Total Accepted:    83.7K
+ * Total Submissions: 169K
+ * Testcase Example:  '[7,2,4,3]\n[5,6,4]'
+ *
+ * You are given two non-empty linked lists representing two non-negative
+ * integers. The most significant digit comes first and each of their nodes
+ * contain a single digit. Add the two numbers and return it as a linked list.
+ * 
+ * You may assume the two numbers do not contain any leading zero, except the
+ * number 0 itself.
+ * 
+ * Follow up:
+ * What if you cannot modify the input lists? In other words, reversing the
+ * lists is not allowed.
+ * 
+ * 
+ * 
+ * Example:
+ * 
+ * Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
+ * Output: 7 -> 8 -> 0 -> 7
+ * 
+ * 
+ */
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} l1
+ * @param {ListNode} l2
+ * @return {ListNode}
+ */
+var addTwoNumbers = function(l1, l2) {
+    const stack1 = [];
+    const stack2 = [];
+    const stack = [];
+
+    let cur1 = l1;
+    let cur2 = l2;
+    let curried = 0;
+
+    while(cur1) {
+        stack1.push(cur1.val);
+        cur1 = cur1.next;
+    }
+
+    while(cur2) {
+        stack2.push(cur2.val);
+        cur2 = cur2.next;
+    }
+
+    let a = null;
+    let b = null;
+
+    while(stack1.length > 0 || stack2.length > 0) {
+        a = Number(stack1.pop()) || 0;
+        b = Number(stack2.pop()) || 0;
+
+        stack.push((a + b + curried) % 10);
+
+        if (a + b + curried >= 10) {
+            curried = 1;
+        } else {
+            curried = 0;
+        }
+    }
+
+    if (curried === 1) {
+        stack.push(1);
+    }
+
+    const dummy = {};
+
+    let current = dummy;
+
+    while(stack.length > 0) {
+        current.next = {
+            val: stack.pop(),
+            next: null
+        }
+
+        current = current.next
+    }
+
+    return dummy.next;
+};
+
+```