Leetcode练习之分治

1. 给表达式加括号

241. Different Ways to Add Parentheses (Medium)

 

class Solution {
    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)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    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;
    }
}