SQL50 基础

1633

  select contest_id,Round(count(*) * 100/(select count(*) from Users),2) as percentage from Users cross join Registerusing(user_id)group by contest_idorder by count(*) desc,contest_id

1211

select query_name , Round(avg(rating/position),2) as quality,
Round(sum(IF(rating<3,1,0))*100/count(*),2) as poor_query_percentage
from Queries 
group by query_name

动态规划基础版

198.打家劫舍

class Solution {public int rob(int[] nums) {int n = nums.length;int dp[] = new int[n];dp[0] = nums[0];if(n>=2)dp[1] = Math.max(nums[0],nums[1]);for(int i=2;i<n;i++){dp[i] = Math.max(dp[i-1],dp[i-2]+nums[i]);}return dp[n-1];}
}

740.删除并获得点数

class Solution {public static int deleteAndEarn(int[] nums) {Map<Integer,Integer> mp = new TreeMap<>();for(int num:nums)mp.put(num,mp.getOrDefault(num,0)+num);int n = mp.size();int dp[][] = new int[n][2];int i = 0;for(Map.Entry<Integer,Integer>entry:mp.entrySet()){dp[i][0] = entry.getKey();dp[i][1] = entry.getValue();i++;}for(int j=1;j<n;j++){if(dp[j][0]-dp[j-1][0]>1)dp[j][1] += dp[j-1][1];else {dp[j][1] = Math.max((j-2>=0?dp[j-2][1]:0)+dp[j][1],dp[j-1][1]);}}return dp[n-1][1];}
}

62.不同路径

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

64.最小路径和

class Solution {public int minPathSum(int[][] grid) {int n = grid.length;int m = grid[0].length;for(int i=1;i<n;i++)grid[i][0] += grid[i-1][0];for(int j=1;j<m;j++)grid[0][j] += grid[0][j-1];for(int i=1;i<n;i++){for(int j=1;j<m;j++)grid[i][j] += Math.min(grid[i-1][j],grid[i][j-1]);}return grid[n-1][m-1];}
}

63.不同路径II

class Solution {public int uniquePathsWithObstacles(int[][] obstacleGrid) {int n = obstacleGrid.length;int m = obstacleGrid[0].length;int dp[][] = new int[n][m];for(int i=0;i<n;i++){if(obstacleGrid[i][0]==1)break;dp[i][0] = 1;}for(int j=0;j<m;j++){if(obstacleGrid[0][j]==1)break;dp[0][j] = 1;}for(int i=1;i<n;i++){for(int j=1;j<m;j++){if(obstacleGrid[i][j]==1)continue;dp[i][j] = dp[i-1][j]+dp[i][j-1];}}return dp[n-1][m-1];}
}