復(fù)習(xí)打卡，學(xué)習(xí)完全背包問題

• 完全背包
• SQL題目

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完全背包

• 518.零錢兌換II；

class Solution {public int change(int amount, int[] coins) {int[] dp = new int[amount + 1];dp[0] = 1;for (int i = 0; i < coins.length; i++) {for (int j = coins[i]; j <= amount; j++) {dp[j] += dp[j - coins[i]];}}return dp[amount];} }

• 377. 組合總和 Ⅳ；

class Solution {public int combinationSum4(int[] nums, int target) {int[] dp = new int[target + 1];dp[0] = 1;for (int i = 0; i <= target; i++) {for (int j = 0; j < nums.length; j++) {if (i >= nums[j]) {dp[i] += dp[i - nums[j]];}}}return dp[target];} }

• 70. 爬樓梯；

class Solution {public int climbStairs(int n) {int[] dp = new int[n + 1];dp[0] = 1;dp[1] = 1;for (int i = 2; i <= n; i++) {for (int j = 1; j <= 2; j++) {dp[i] += dp[i - j];}}return dp[n];} }

• 322. 零錢兌換；

class Solution {public int coinChange(int[] coins, int amount) {int[] dp = new int[amount + 1];Arrays.fill(dp, Integer.MAX_VALUE);dp[0] = 0;for (int i = 0; i < coins.length; i++) {for (int j = coins[i]; j <= amount; j++) {if (dp[j - coins[i]] != Integer.MAX_VALUE) {dp[j] = Math.min(dp[j], dp[j - coins[i]] + 1);}}}return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];} }

SQL題目

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• 查詢課程編號為“01”的課程比“02”的課程成績高的所有學(xué)生的學(xué)號

-- 查詢課程編號為“01”的課程比“02”的課程成績高的所有學(xué)生的學(xué)號 select s_id, s_score from score where c_id = '01'; select s_id, s_score from score where c_id = '02';select a.s_id from (select s_id, s_score from score where c_id = '01') ainner join (select s_id, s_score from score where c_id = '02') b on a.s_id = b.s_id where a.s_score > b.s_score;

• 查詢平均成績大于60分的學(xué)生的學(xué)號和平均成績

-- 查詢平均成績大于60分的學(xué)生的學(xué)號和平均成績 select s_id,avg(s_score) from score group by s_id having avg(s_score) > 60;

• 查詢所有學(xué)生的學(xué)號、姓名、選課數(shù)、總成績

-- 查詢所有學(xué)生的學(xué)號、姓名、選課數(shù)、總成績 select student.s_id, s_name, count(c_id), sum(case when s_score is null then 0 else s_score end) from studentleft join score s on student.s_id = s.s_id group by student.s_id, s_name;

• 查詢沒學(xué)過張三老師課程的學(xué)生的學(xué)號和姓名;

-- 查詢沒學(xué)過張三老師課程的學(xué)生的學(xué)號和姓名;select student.s_id from studentinner join score s on student.s_id = s.s_idinner join course c on s.c_id = c.c_idinner join teacher t on c.t_id = t.t_id where t_name = '張三';select s_id, s_name from student where s_id not in (select student.s_idfrom studentinner join score s on student.s_id = s.s_idinner join course c on s.c_id = c.c_idinner join teacher t on c.t_id = t.t_idwhere t_name = '張三');

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