113 lines
3.0 KiB
Java
113 lines
3.0 KiB
Java
# Leetcode 题解 - 分治
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<!-- GFM-TOC -->
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* [Leetcode 题解 - 分治](#leetcode-题解---分治)
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* [1. 给表达式加括号](#1-给表达式加括号)
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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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<!-- GFM-TOC -->
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## 1. 给表达式加括号
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241\. Different Ways to Add Parentheses (Medium)
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[Leetcode](https://leetcode.com/problems/different-ways-to-add-parentheses/description/) / [力扣](https://leetcode-cn.com/problems/different-ways-to-add-parentheses/description/)
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```html
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Input: "2-1-1".
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((2-1)-1) = 0
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(2-(1-1)) = 2
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Output : [0, 2]
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```
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```java
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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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}
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```
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## 2. 不同的二叉搜索树
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95\. Unique Binary Search Trees II (Medium)
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[Leetcode](https://leetcode.com/problems/unique-binary-search-trees-ii/description/) / [力扣](https://leetcode-cn.com/problems/unique-binary-search-trees-ii/description/)
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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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```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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if (n < 1) {
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return new LinkedList<TreeNode>();
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}
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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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