博客
关于我
Weekly Contest 133
阅读量:426 次
发布时间:2019-03-06

本文共 7511 字,大约阅读时间需要 25 分钟。

1030. Matrix Cells in Distance Order

We are given a matrix with R rows and C columns has cells with integer coordinates (r, c), where 0 <= r < R and 0 <= c < C.

Additionally, we are given a cell in that matrix with coordinates (r0, c0).

Return the coordinates of all cells in the matrix, sorted by their distance from (r0, c0) from smallest distance to largest distance.  Here, the distance between two cells (r1, c1) and (r2, c2) is the Manhattan distance, |r1 - r2| + |c1 - c2|.  (You may return the answer in any order that satisfies this condition.)

 

Example 1:

Input: R = 1, C = 2, r0 = 0, c0 = 0Output: [[0,0],[0,1]]Explanation: The distances from (r0, c0) to other cells are: [0,1]

Example 2:

Input: R = 2, C = 2, r0 = 0, c0 = 1Output: [[0,1],[0,0],[1,1],[1,0]]Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2]The answer [[0,1],[1,1],[0,0],[1,0]] would also be accepted as correct.

Example 3:

Input: R = 2, C = 3, r0 = 1, c0 = 2Output: [[1,2],[0,2],[1,1],[0,1],[1,0],[0,0]]Explanation: The distances from (r0, c0) to other cells are: [0,1,1,2,2,3]There are other answers that would also be accepted as correct, such as [[1,2],[1,1],[0,2],[1,0],[0,1],[0,0]].

 

Note:

  1. 1 <= R <= 100
  2. 1 <= C <= 100
  3. 0 <= r0 < R
  4. 0 <= c0 < C

 

Approach #1: 

class Solution {public:    vector
> allCellsDistOrder(int R, int C, int r0, int c0) { map
>> m; for (int i = 0; i < R; ++i) { for (int j = 0; j < C; ++j) { int dis = abs(i - r0) + abs(j - c0); m[dis].push_back({i, j}); } } int index = 0; vector
> ret = vector(R*C, vector
(2)); map
>>::iterator it; for (it = m.begin(); it != m.end(); ++it) { vector
> temp = it->second; for (int i = 0; i < temp.size(); ++i) { ret[index][0] = temp[i].first; ret[index][1] = temp[i].second; index++; } } return ret; }};

  

1029. Two City Scheduling

There are 2N people a company is planning to interview. The cost of flying the i-th person to city A is costs[i][0], and the cost of flying the i-th person to city B is costs[i][1].

Return the minimum cost to fly every person to a city such that exactly N people arrive in each city.

 

Example 1:

Input: [[10,20],[30,200],[400,50],[30,20]]Output: 110Explanation: The first person goes to city A for a cost of 10.The second person goes to city A for a cost of 30.The third person goes to city B for a cost of 50.The fourth person goes to city B for a cost of 20.The total minimum cost is 10 + 30 + 50 + 20 = 110 to have half the people interviewing in each city.

 

Note:

  1. 1 <= costs.length <= 100
  2. It is guaranteed that costs.length is even.
  3. 1 <= costs[i][0], costs[i][1] <= 1000

 

Approach #1: 

class Solution {    public int twoCitySchedCost(int[][] costs) {        int N = costs.length / 2;        int[][] dp = new int[N+1][N+1];        for (int i = 1; i <= N; ++i) {            dp[i][0] = dp[i-1][0] + costs[i-1][0];        }        for (int i = 1; i <= N; ++i) {            dp[0][i] = dp[0][i-1] + costs[i-1][1];        }        for (int i = 1; i <= N; ++i) {            for (int j = 1; j <= N; ++j) {                dp[i][j] = Math.min(dp[i-1][j] + costs[i+j-1][0], dp[i][j-1] + costs[i+j-1][1]);            }        }        return dp[N][N];    }}

  

1031. Maximum Sum of Two Non-Overlapping Subarrays

Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M.  (For clarification, the L-length subarray could occur before or after the M-length subarray.)

Formally, return the largest V for which V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) and either:

  • 0 <= i < i + L - 1 < j < j + M - 1 < A.length, or
  • 0 <= j < j + M - 1 < i < i + L - 1 < A.length.

 

Example 1:

Input: A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2Output: 20Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.

Example 2:

Input: A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2Output: 29Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.

Example 3:

Input: A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3Output: 31Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3.

 

Note:

  1. L >= 1
  2. M >= 1
  3. L + M <= A.length <= 1000
  4. 0 <= A[i] <= 1000

 

Approach #1:

class Solution {    public int maxSumTwoNoOverlap(int[] A, int L, int M) {        for (int i = 1; i < A.length; ++i) {            A[i] += A[i-1];        }        int res = A[L+M-1], Lmax = A[L-1], Mmax = A[M-1];        for (int i = L + M; i < A.length; ++i) {            Lmax = Math.max(Lmax, A[i-M] - A[i-M-L]);            Mmax = Math.max(Mmax, A[i-L] - A[i-M-L]);            res = Math.max(res, Math.max(Lmax + A[i] - A[i-M], Mmax + A[i] - A[i-L]));        }                return res;    }}

  

1032. Stream of Characters

Implement the StreamChecker class as follows:

  • StreamChecker(words): Constructor, init the data structure with the given words.
  • query(letter): returns true if and only if for some k >= 1, the last k characters queried (in order from oldest to newest, including this letter just queried) spell one of the words in the given list.

 

Example:

StreamChecker streamChecker = new StreamChecker(["cd","f","kl"]); // init the dictionary.streamChecker.query('a');          // return falsestreamChecker.query('b');          // return falsestreamChecker.query('c');          // return falsestreamChecker.query('d');          // return true, because 'cd' is in the wordliststreamChecker.query('e');          // return falsestreamChecker.query('f');          // return true, because 'f' is in the wordliststreamChecker.query('g');          // return falsestreamChecker.query('h');          // return falsestreamChecker.query('i');          // return falsestreamChecker.query('j');          // return falsestreamChecker.query('k');          // return falsestreamChecker.query('l');          // return true, because 'kl' is in the wordlist

 

Note:

  • 1 <= words.length <= 2000
  • 1 <= words[i].length <= 2000
  • Words will only consist of lowercase English letters.
  • Queries will only consist of lowercase English letters.
  • The number of queries is at most 40000.

 

Approach #1:

class StreamChecker {    public class TriNode {        boolean isEnd = false;        TriNode[] next = new TriNode[26];    }        TriNode root = new TriNode();    StringBuilder buf = new StringBuilder();        void insert(String word) {        TriNode temp = root;        for (int i = 0; i < word.length(); ++i) {            char ch = word.charAt(word.length()-i-1);            if (temp.next[ch-'a'] == null) temp.next[ch-'a'] = new TriNode();            temp = temp.next[ch-'a'];        }        temp.isEnd = true;    }        public StreamChecker(String[] words) {        for (String word : words) {            insert(word);        }    }        public boolean query(char letter) {        buf.append(letter);        // System.out.println(buf.toString());        TriNode p = root;        for (int i = buf.length() - 1; i >= 0; --i) {            char ch = buf.charAt(i);            p = p.next[ch-'a'];            if (p == null) return false;            if (p.isEnd) return true;        }        return false;    }}/** * Your StreamChecker object will be instantiated and called as such: * StreamChecker obj = new StreamChecker(words); * boolean param_1 = obj.query(letter); */

  

 

转载地址:http://iftuz.baihongyu.com/

你可能感兴趣的文章
mysql更新频率_MySQL优化之如何了解SQL的执行频率
查看>>
mysql替换表的字段里面内容
查看>>
MySQL最多能有多少连接
查看>>
MySQL最大建议行数 2000w,靠谱吗?
查看>>
MySQL有哪些锁
查看>>
MySQL服务器安装(Linux)
查看>>
mysql服务器查询慢原因分析方法
查看>>
mysql服务无法启动的问题
查看>>
MySQL杂谈
查看>>
mysql权限
查看>>
mysql条件查询
查看>>
MySQL条件查询
查看>>
MySQL架构与SQL的执行流程_1
查看>>
MySQL架构与SQL的执行流程_2
查看>>
MySQL架构介绍
查看>>
MySQL架构优化
查看>>
mysql架构简介、及linux版的安装
查看>>
MySQL查看数据库相关信息
查看>>
MySQL查看表结构和表中数据
查看>>
MySQL查询优化:LIMIT 1避免全表扫描
查看>>