力扣題【簡單級別】1

2019.9.10

771.寶石與石頭

public int numJewelsInStones(String J, String S) {
        if (J == null || S == null) return 0;
        if (J.length() == 0 || S.length() == 0) return 0;
        Byte[] b = new Byte[58];
        int count = 0;
        for(char ch: J.toCharArray()){
            b[ch - 'A'] = 1;//'A'的ASCII碼是65，'A'和'z'的ASCII碼相差58
        }
        for(char ch: S.toCharArray()){
            if(b[ch - 'A'] != null && b[ch - 'A']== 1)
                count++;
        }
        return count;
    }

2019.9.17

1108.IP 地址無效化

和劍指offer的【字符串-替換空格】差不多的思路

public String defangIPaddr(String address) {
        int count = 0;//記錄字符串中.的個數
        for(int i = 0; i < address.length(); i++){
            if(address.charAt(i) == '.')
                count++;
        }
        int oldLength = address.length();
        StringBuffer s = new StringBuffer(address);
        s.setLength(oldLength+count*2);
        for(int k = oldLength - 1; k > -1; k--){
            if(count > 0){
                if(s.charAt(k) != '.'){
                    s.setCharAt(k+count*2,s.charAt(k));
                }else{
                    s.replace(k+count*2-2,k+count*2+1,"[.]");
                    count--;
                }
            }
            else
                break;
        }
        return s.toString();
    }

237.刪除鏈表中的節點

這題題意有點難理解，deleteNode傳參是待刪除的那個節點

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public void deleteNode(ListNode node) {
        node.val = node.next.val;
        node.next = node.next.next;
        node = null;
        //在鏈表上已刪除節點後，清空它爲null，方便GC標記回收
    }
}

182.查找重複的電子郵箱（原來力扣還有MySQL的題）

count 是聚合函數，使用 count(*) 要比count(字段) 要快，省略了判斷字段是否爲 null 的過程，mysql 對 count(*) 做了優化。

SELECT Email FROM Person GROUP BY Email HAVING COUNT(*) > 1;
# 或者
# select distinct a.Email from Person a, Person b where a.Email = b.Email and a.Id != b.Id 

938.二叉搜索樹的範圍和（又是題目都看不懂的我，還有救嗎）

參數給了二叉搜索樹所屬的兩個節點，求二叉搜索樹節點值按序排列時兩節點值間（包括兩節點）的所有節點值的和。可以遍歷節點去判斷節點值是否在這兩個節點值間，在配合剪枝減少沒必要的節點遍歷

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.Stack;
class Solution {
    private int res = 0;//記錄和
    public int rangeSumBST(TreeNode root, int L, int R) {
        if (root == null) return 0;
        else{
            dfs(root, L, R);
        }
        return res;            
    }
    
    //遞歸
    private void dfs(TreeNode root, int L, int R){
        if(root != null){
            if(root.val > L){
                dfs(root.left, L, R);
            }
            if(root.val < R){
                dfs(root.right, L, R);
            }
            if(root.val >= L && root.val <= R){
                res += root.val;
            }
        }
    }
    
    //迭代
    private void dfs1(TreeNode root, int L, int R){
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode t = stack.pop();
            if(t != null){
                if(t.val > L){
                    stack.push(t.left);
                }
                if(t.val < L){
                    stack.push(t.right);
                }
                if(t.val >= L && t.val <= R){
                    res += t.val;
                }
            }
        }
    }
}

2019.9.18

1021.刪除最外層的括號

//用棧記錄左右括號的匹配
public static String removeOuterParentheses(String S) {
    Stack<Character> stack = new Stack<>();
    StringBuffer sB = new StringBuffer();
    int start = 0;
    for(int i = 0; i < S.length(); i++){
        if(S.charAt(i) == '('){
            stack.push(S.charAt(i));
        }else {
            stack.pop();
            if(stack.empty()){
                sB.append(S.substring(start+1, i));
                start = i + 1;
            }
        }
    }
    return  sB.toString();
}

//直接計數，效率更高。
//最外面的左括號跳過，L初始爲1；當右括號數量等於左括號數量的位置就是最外面的右括號
public static String removeOuterParentheses1(String S) {
    StringBuffer sB = new StringBuffer();
    int L = 1, R = 0;
    for(int i = 1; i < S.length(); i++){
        if(S.charAt(i) == '('){
            sB.append(S.charAt(i));
            L++;
        }else {
            R++;
            if(L == R){
                i++;
                L = 1;
                R = 0;
            }else {
                sB.append(S.charAt(i));
            }
        }
    }
    return sB.toString();
}

595.大的國家（拯救信心繫列）

SELECT name, population, area FROM WORLD WHERE area > 3000000 OR population > 25000000

709.轉換成小寫字母（拯救信心繫列）

 public String toLowerCase(String str) {
        StringBuffer sB = new StringBuffer(str);
        for(int i = 0; i < str.length(); i++){
            int a = str.charAt(i) - 'A';
            if(a >= 0 && a <= 25){
                sB.setCharAt(i, (char)('a'+a));
            }
        }
        return sB.toString();
    }

    //效率更高。ASCII碼 'A'=65 'a'=97
    public String toLowerCase1(String str) {
        StringBuffer sB = new StringBuffer(str);
        for(int i = 0; i < str.length(); i++){
            if(str.charAt(i) >= 65 && str.charAt(i) <= 90){
                sB.setCharAt(i, (char)(str.charAt(i)+32));
            }
        }
        return sB.toString();
    }

617.合併二叉樹（明明想法一樣，我就遞歸不出來，哭。。。）

public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }
    }
    public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
        if(t1 == null) return t2;
        if(t2 == null) return t1;
        t1.val = t1.val + t2.val;
        t1.left = mergeTrees(t1.left, t2.left);
        t1.right = mergeTrees(t1.right, t2.right);
        return t1;
    }

804.唯一摩爾斯密碼詞（拯救信心繫列）

改進：String+HashMap ——> StringBuffer+HashSet

	public int uniqueMorseRepresentations(String[] words) {
        String[] array = {".-","-...","-.-.","-..",".","..-.","--.","....","..",".---","-.-",".-..","--","-.","---",".--.","--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
        //HashMap<String, Byte> map = new HashMap<>();
        HashSet<String> set = new HashSet<>();//考慮用
        for(String s: words){
            //要操作字符串的話，StringBuffer比String高效一點
            StringBuffer tmp = new StringBuffer();
            for(int i = 0; i < s.length(); i++){
                tmp.append(array[s.charAt(i)-'a']);
            }
            //不用判斷map是否包含key，因爲如果key已存在，put會把value直接覆蓋
            //map.put(tmp.toString(),(byte)0);
            set.add(tmp.toString());
        }
        //return map.size();
        return set.size();
    }

832.翻轉圖像（拯救信心繫列）

public static int[][] flipAndInvertImage(int[][] A) {
    int col = A[0].length;
    for(int i = 0; i < A.length; i++){
        for(int k = 0; k < (col+1)/2; k++){
            if(A[i][k] == A[i][col-1-k]){
                if(k != col-1-k){
                    A[i][col-1-k]= (A[i][k]==1?0:1);
                }
                A[i][k] = (A[i][k]==1?0:1);
            }
        }
    }
    return A;
}

620.有趣的電影（。。第一次竟然沒敲出來）

MySQL用mod(row_name, number)函數來取模；用id&1來判斷奇數更快；根據最左前綴和and的短路，讓where先判斷id再去判斷description，更快；ANSI標準中是用<>表示不等於，雖然也可以用!=

SELECT * FROM cinema WHERE id&1 AND description <> 'boring'  ORDER BY rating DESC

461.漢明距離

先異或，然後用劍指offer裏的【二進制數中1的個數】

public int hammingDistance(int x, int y) {
        int xor = x ^ y;
        int count = 0;
        while(xor != 0){
            count += xor&1;
            xor>>=1;
        }
        return count;
    }

226.翻轉二叉樹（劍指offer【二叉樹的鏡像】）

private TreeNode tmp = null;
public TreeNode invertTree(TreeNode root) {
    if(root == null) return null;
    tmp = root.left;
    root.left = root.right;
    root.right = tmp;
    invertTree(root.left);
    invertTree(root.right);
    return root;
}

657.機器人能否返回原點

public boolean judgeCircle(String moves) {
        int x = 0, y = 0;
        for(int i = 0; i < moves.length(); i++){
            switch (moves.charAt(i)){
                case 'L' : x++; break;
                case 'R' : x--; break;
                case 'U' : y++; break;
                case 'D' : y--; break; 
            }
        }
         return x == 0 && y == 0;
    }

1051.高度檢查器

public static int heightChecker(int[] heights) {
    int[] res = new int[100];//用於計數排序
    int count = 0;
    for(int i = 0; i < heights.length; i++){
        res[heights[i]-1]++;
    }
    int k = 0;
    for(int i = 0; i < heights.length; i++){
        while (res[k] == 0){
            k++;
        }
        if(k!=heights[i]-1) count++;
        res[k]--;
    }
    return count;
}

627.交換工資

抄的。SQL有char(int)、ascii(char)函數？牛逼！！

update salary set sex = char(ASCII(sex) ^ ASCII('m') ^ ASCII('f')); #暫時這個最棒
#UPDATE salary SET sex = char(ascii('m') + ascii('f') - ascii(sex));

官方的case…when…也可以學學

UPDATE salary
SET sex = 
	CASE sex
        WHEN 'm' THEN 'f'
        ELSE 'm'
    END;

咱變着花樣來，還有if

update salary set sex=if(sex = 'f', 'm','f');