Leetcode题解——算法思想之分治-优快云博客

1. 给表达式加括号

241. Different Ways to Add Parentheses (Medium)

Input: "2-1-1".

((2-1)-1) = 0
(2-(1-1)) = 2

Output : [0, 2]

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)

给定一个数字 n，要求生成所有值为 1...n 的二叉搜索树。

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

public List<TreeNode> generateTrees(int n) {
    if (n < 1) { return new LinkedList<TreeNode>(); } 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; }