2019-04-21 10:36:08 +08:00
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<!-- GFM-TOC -->
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2019-03-27 20:57:37 +08:00
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* [1. 给表达式加括号](#1-给表达式加括号)
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2019-04-20 13:14:16 +08:00
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* [2. 不同的二叉搜索树](#2-不同的二叉搜索树)
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2019-04-21 10:36:08 +08:00
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<!-- GFM-TOC -->
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2019-03-08 20:31:07 +08:00
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2019-03-27 20:57:37 +08:00
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# 1. 给表达式加括号
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[241. Different Ways to Add Parentheses (Medium)](https://leetcode.com/problems/different-ways-to-add-parentheses/description/)
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2019-03-08 20:31:07 +08:00
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```html
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2019-03-27 20:57:37 +08:00
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Input: "2-1-1".
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2019-03-08 20:31:07 +08:00
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2019-03-27 20:57:37 +08:00
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((2-1)-1) = 0
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(2-(1-1)) = 2
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2019-03-08 20:31:07 +08:00
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2019-03-27 20:57:37 +08:00
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Output : [0, 2]
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2019-03-08 20:31:07 +08:00
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```
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```java
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2019-03-27 20:57:37 +08:00
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public List<Integer> diffWaysToCompute(String input) {
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List<Integer> ways = new ArrayList<>();
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for (int i = 0; i < input.length(); i++) {
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char c = input.charAt(i);
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if (c == '+' || c == '-' || c == '*') {
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List<Integer> left = diffWaysToCompute(input.substring(0, i));
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List<Integer> right = diffWaysToCompute(input.substring(i + 1));
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for (int l : left) {
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for (int r : right) {
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switch (c) {
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case '+':
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ways.add(l + r);
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break;
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case '-':
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ways.add(l - r);
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break;
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case '*':
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ways.add(l * r);
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break;
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}
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}
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}
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}
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}
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if (ways.size() == 0) {
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ways.add(Integer.valueOf(input));
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}
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return ways;
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2019-03-08 20:31:07 +08:00
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}
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```
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2019-03-27 20:57:37 +08:00
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2019-04-20 13:14:16 +08:00
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# 2. 不同的二叉搜索树
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2019-04-15 23:51:41 +08:00
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[95. Unique Binary Search Trees II (Medium)](https://leetcode.com/problems/unique-binary-search-trees-ii/description/)
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2019-04-20 13:14:16 +08:00
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给定一个数字 n,要求生成所有值为 1...n 的二叉搜索树。
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2019-04-15 23:51:41 +08:00
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```html
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Input: 3
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Output:
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[
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[1,null,3,2],
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[3,2,null,1],
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[3,1,null,null,2],
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[2,1,3],
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[1,null,2,null,3]
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]
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Explanation:
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The above output corresponds to the 5 unique BST's shown below:
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1 3 3 2 1
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\ / / / \ \
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3 2 1 1 3 2
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/ / \ \
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2 1 2 3
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```
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```java
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public List<TreeNode> generateTrees(int n) {
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2019-05-14 17:05:21 +08:00
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if (n < 1) {
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return new LinkedList<TreeNode>();
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}
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2019-04-15 23:51:41 +08:00
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return generateSubtrees(1, n);
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}
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private List<TreeNode> generateSubtrees(int s, int e) {
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List<TreeNode> res = new LinkedList<TreeNode>();
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if (s > e) {
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res.add(null);
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return res;
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}
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for (int i = s; i <= e; ++i) {
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List<TreeNode> leftSubtrees = generateSubtrees(s, i - 1);
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List<TreeNode> rightSubtrees = generateSubtrees(i + 1, e);
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for (TreeNode left : leftSubtrees) {
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for (TreeNode right : rightSubtrees) {
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TreeNode root = new TreeNode(i);
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root.left = left;
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root.right = right;
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res.add(root);
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}
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}
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}
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return res;
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}
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```
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2019-03-27 20:57:37 +08:00
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2019-04-20 13:14:16 +08:00
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2019-06-13 08:50:30 +08:00
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# 与我交流
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2019-06-10 11:23:18 +08:00
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2019-06-13 08:50:30 +08:00
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😄 你可以在我的微信公众号后台与我交流。✔️ 另外,公众号提供了该项目的离线阅读版本,后台回复 "下载" 即可领取。🧾 也提供了一份技术面试复习大纲,不仅系统整理了面试知识点,而且标注了各个知识点的重要程度,从而帮你理清多而杂的面试知识点,后台回复 "大纲" 即可领取。我基本是按照这个大纲来进行复习的,对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习,就不用看很多不重要的内容,也可以知道哪些内容很重要从而多安排一些复习时间。
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2019-06-10 11:23:18 +08:00
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2019-06-12 12:26:39 +08:00
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