为运算表达式设计优先级

力扣

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class Solution {
    public List<Integer> diffWaysToCompute(String input) {
        List<Integer> result = new ArrayList<>();
        //终止条件
        if (!input.contains("+") && !input.contains("-") && !input.contains("*")) {
            result.add(Integer.valueOf(input));
            return result;
        }
        for (int i = 0; i < input.length(); i++) {
            if (input.charAt(i) == '+' || input.charAt(i) == '-' || input.charAt(i) == '*') {
                for (Integer left : diffWaysToCompute(input.substring(0, i))) {
                    for (Integer right : diffWaysToCompute(input.substring(i + 1))) {
                        if (input.charAt(i) == '+') {
                            result.add(left + right);
                        } else if (input.charAt(i) == '-') {
                            result.add(left - right);
                        } else if (input.charAt(i) == '*') {
                            result.add(left * right);
                        }
                    }
                }
            }
        }
        return result;
    }
}

不同的二叉搜索树Ⅱ

力扣

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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>();
        //终止条件
        if (s > e) {
            res.add(null);
            return res;
        }
        for (int i = s; i <= e; i++) {
            // i作为根节点
            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;
    }
}