最长公共子序列-II&&在两个长度相等的排序数组中找到上中位数&&判断一棵二叉树是否为搜索树和完全二叉树

NC92 最长公共子序列-II

题目链接

1、解题思路

使用dp的思想求出最长长度，然后根据长度反推出字符串。

2、代码

import java.util.*;


public class Solution {
    /**
     * longest common subsequence
     * @param s1 string字符串 the string
     * @param s2 string字符串 the string
     * @return string字符串
     */
    public String LCS (String s1, String s2) {
        // write code here
        // write code here
        int r = s1.length();
        int c = s2.length();
        int[][] dp = new int[r + 1][c + 1];
        for (int i = 0; i <= c; i++) {
            dp[0][i] = 0;
        }
        for (int i = 0; i <= r; i++) {
            dp[i][0] = 0;
        }
        for (int i = 1; i <= r; i++) {
            for (int j = 1; j <= c; j++) {
                char c1 = s1.charAt(i-1);
                char c2 = s2.charAt(j-1);
                if (c1 == c2) {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                } else {
                    dp[i][j] = dp[i - 1][j] > dp[i][j - 1] ? dp[i - 1][j] : dp[i][j - 1];
                }
            }
        }
        if (dp[r][c] == 0) {
            return "-1";
        }
        char[] arr = new char[dp[r][c]];
        int index = arr.length - 1;
        while (index >= 0) {
            if (s1.charAt(r-1) == s2.charAt(c-1)) {
                arr[index--] = s1.charAt(r-1);
                r--;
                c--;
            } else if (dp[r][c] == dp[r - 1][c]) {
                r--;
            } else {
                c--;
            }
        }
        return new String(arr);
    }
}

NC36 在两个长度相等的排序数组中找到上中位数

题目链接

1、解题思路

双指针跑一遍就好了。

2、代码

import java.util.*;


public class Solution {
    /**
     * find median in two sorted array
     * @param arr1 int整型一维数组 the array1
     * @param arr2 int整型一维数组 the array2
     * @return int整型
     */
    public int findMedianinTwoSortedAray (int[] arr1, int[] arr2) {
        // write code here
        int l1 = 0;
        int l2 = 0;
        int index = (arr1.length + arr2.length) / 2 - 1;
        for (int i = 0; i < index; i++) {
            if (arr1[l1] <= arr2[l2]) {
                l1++;
            } else {
                l2++;
            }
        }
        return arr1[l1] < arr2[l2] ? arr1[l1] : arr2[l2];
    }
}

NC60 判断一棵二叉树是否为搜索树和完全二叉树

题目链接

1、解题思路

搜索树：可以先序判断数组是否有序

完全二叉树：层次遍历，出现空节点后面不能再出现节点。

2、代码

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param root TreeNode类 the root
     * @return bool布尔型一维数组
     */
    private void preOrder(TreeNode root, List<Integer> list) {
        if (root == null) {
            return;
        }
        preOrder(root.left, list);
        list.add(root.val);
        preOrder(root.right, list);
    }

    private boolean judge1(TreeNode root) {
        if (root == null) {
            return true;
        }
        List<Integer> list = new ArrayList<>();
        preOrder(root, list);
        for (int i = 1; i < list.size(); i++) {
            if (list.get(i) < list.get(i - 1)) {
                return false;
            }
        }
        return true;
    }

    private boolean judge2(TreeNode node) {
        if (node == null) {
            return true;
        }
        boolean f = false;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(node);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeNode temp = queue.poll();
                if (temp.left != null) {
                    if (f) {
                        return false;
                    }
                    queue.add(temp.left);
                } else {
                    f = true;
                }
                if (temp.right != null) {
                    if (f) {
                        return false;
                    }
                    queue.add(temp.right);
                } else {
                    f = true;
                }
            }
        }
        return true;
    }

    public boolean[] judgeIt(TreeNode root) {
        // write code here
        boolean[] judge = new boolean[2];
        judge[0] = judge1(root);
        judge[1] = judge2(root);
        return judge;
    }
}