劍指Offer之Java演演算法習題精講N叉樹的遍歷及陣列與字串

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;
    public Node() {}
    public Node(int _val) {
        val = _val;
    }
    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
 
class Solution {
    ArrayList<Integer> list = new ArrayList<Integer>();
    public List<Integer> preorder(Node root) {
        if (root == null) return list;
        list.add(root.val);
        for(int i=0;i<root.children.size();i++){
            preorder(root.children.get(i));
        }
        return list;
    }
}

/*
// Definition for a Node.
class Node {
    public int val;
    public List<Node> children;
    public Node() {}
    public Node(int _val) {
        val = _val;
    }
    public Node(int _val, List<Node> _children) {
        val = _val;
        children = _children;
    }
};
*/
 
class Solution {
    ArrayList<Integer> list = new ArrayList<Integer>();
    public List<Integer> postorder(Node root) {
        if(root==null) return list;
        for(int i = 0;i<root.children.size();i++){
            postorder(root.children.get(i));
        }
        list.add(root.val);
        return list;
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public String tree2str(TreeNode root) {
        // root為空返回""
        if(root==null){
            return "";
        }
        // root左右為空
        if(root.left==null&&root.right==null){
            return root.val+"";
        }
        // root右為空
        if(root.right==null){
            return root.val+"("+tree2str(root.left)+")";
        }
        // root左右不空或是左不為空
        return root.val+"("+tree2str(root.left)+")"+"("+tree2str(root.right)+")";
    }
}

class Solution {
    public int maximumProduct(int[] nums) {
        Arrays.sort(nums);
        int n = nums.length;
        // 拿到最大三個數，或者最小兩個數和最大一個數
        int ans = Math.max(nums[n-1]*nums[n-2]*nums[n-3],nums[n-1]*nums[0]*nums[1]);
        return ans;
    }
}
 
// 拿到最大三個數，或者最小兩個數和最大一個數
class Solution {
    public int maximumProduct(int[] nums) {
        // 最小的和第二小的
        int min1 = Integer.MAX_VALUE, min2 = Integer.MAX_VALUE;
        // 最大的、第二大的和第三大的
        int max1 = Integer.MIN_VALUE, max2 = Integer.MIN_VALUE, max3 = Integer.MIN_VALUE;
        for(int n:nums){
            if(n<min1){
                min2 = min1;
                min1 = n;
            }else if(n<min2){
                min2 = n;
            }
            if(n>max1){
                max3 = max2;
                max2 = max1;
                max1 = n;
            }else if(n>max2){
                max3 = max2;
                max2 = n;
            }else if(n>max3){
                max3 = n;
            }
        }
        return Math.max(min1*min2*max1,max1*max2*max3);
    }
}