110 lines
3.0 KiB
Markdown
110 lines
3.0 KiB
Markdown
<!-- GFM-TOC -->
|
||
* [1. 给表达式加括号](#1-给表达式加括号)
|
||
* [2. 不同的二叉搜索树](#2-不同的二叉搜索树)
|
||
<!-- GFM-TOC -->
|
||
|
||
|
||
# 1. 给表达式加括号
|
||
|
||
[241. Different Ways to Add Parentheses (Medium)](https://leetcode.com/problems/different-ways-to-add-parentheses/description/)
|
||
|
||
```html
|
||
Input: "2-1-1".
|
||
|
||
((2-1)-1) = 0
|
||
(2-(1-1)) = 2
|
||
|
||
Output : [0, 2]
|
||
```
|
||
|
||
```java
|
||
public List<Integer> diffWaysToCompute(String input) {
|
||
List<Integer> ways = new ArrayList<>();
|
||
for (int i = 0; i < input.length(); i++) {
|
||
char c = input.charAt(i);
|
||
if (c == '+' || c == '-' || c == '*') {
|
||
List<Integer> left = diffWaysToCompute(input.substring(0, i));
|
||
List<Integer> right = diffWaysToCompute(input.substring(i + 1));
|
||
for (int l : left) {
|
||
for (int r : right) {
|
||
switch (c) {
|
||
case '+':
|
||
ways.add(l + r);
|
||
break;
|
||
case '-':
|
||
ways.add(l - r);
|
||
break;
|
||
case '*':
|
||
ways.add(l * r);
|
||
break;
|
||
}
|
||
}
|
||
}
|
||
}
|
||
}
|
||
if (ways.size() == 0) {
|
||
ways.add(Integer.valueOf(input));
|
||
}
|
||
return ways;
|
||
}
|
||
```
|
||
|
||
# 2. 不同的二叉搜索树
|
||
|
||
[95. Unique Binary Search Trees II (Medium)](https://leetcode.com/problems/unique-binary-search-trees-ii/description/)
|
||
|
||
给定一个数字 n，要求生成所有值为 1...n 的二叉搜索树。
|
||
|
||
```html
|
||
Input: 3
|
||
Output:
|
||
[
|
||
[1,null,3,2],
|
||
[3,2,null,1],
|
||
[3,1,null,null,2],
|
||
[2,1,3],
|
||
[1,null,2,null,3]
|
||
]
|
||
Explanation:
|
||
The above output corresponds to the 5 unique BST's shown below:
|
||
|
||
1 3 3 2 1
|
||
\ / / / \ \
|
||
3 2 1 1 3 2
|
||
/ / \ \
|
||
2 1 2 3
|
||
```
|
||
|
||
```java
|
||
public List<TreeNode> generateTrees(int n) {
|
||
return generateSubtrees(1, n);
|
||
}
|
||
|
||
private List<TreeNode> generateSubtrees(int s, int e) {
|
||
List<TreeNode> res = new LinkedList<TreeNode>();
|
||
if (s > e) {
|
||
res.add(null);
|
||
return res;
|
||
}
|
||
for (int i = s; i <= e; ++i) {
|
||
List<TreeNode> leftSubtrees = generateSubtrees(s, i - 1);
|
||
List<TreeNode> rightSubtrees = generateSubtrees(i + 1, e);
|
||
for (TreeNode left : leftSubtrees) {
|
||
for (TreeNode right : rightSubtrees) {
|
||
TreeNode root = new TreeNode(i);
|
||
root.left = left;
|
||
root.right = right;
|
||
res.add(root);
|
||
}
|
||
}
|
||
}
|
||
return res;
|
||
}
|
||
```
|
||
|
||
|
||
|
||
|
||
</br><div align="center">🎨 关注公众号 CyC2018 获取更多精彩内容！在公众号后台回复关键字 **资料** 可领取一份技术面试复习思维导图，帮你理清多而杂的面试知识点。
|
||
<div align="center"><img width="180px" src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7.jpg"></img></div>
|