LeetCode 947. Most Stones Removed with Same Row or Column
原题链接在这里:https://leetcode.com/problems/most-stones-removed-with-same-row-or-column/
题目:
On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may have at most one stone.
Now, a move consists of removing a stone that shares a column or row with another stone on the grid.
What is the largest possible number of moves we can make?
Example 1:
Input: stones = [[0,0],[0,1],[1,0],[1,2],[2,1],[2,2]]
Output: 5
Example 2:
Input: stones = [[0,0],[0,2],[1,1],[2,0],[2,2]]
Output: 3
Example 3:
Input: stones = [[0,0]]
Output: 0
Note:
1 <= stones.length <= 1000
0 <= stones[i][j] < 10000
题解:
The number of most stones removed = stones count - unions count
If a stone in on a row or column same as other stone, then two stones are in the same union.
Use two HashMap to maintain row and column to stone index relationship. If the key already exist, then there is previous stone with same row or column existing already.
Find parents of two stones, if they are not equal, then union.
Time Complexity: O(nlogn). find takes O(logn). Since use path compression and union by size, amortize O(1).
Space: O(n).
AC Java:
class Solution {
Map<Integer, Integer> row = new HashMap<>();
Map<Integer, Integer> colum = new HashMap<>();
int [] parent;
int [] size; public int removeStones(int[][] stones) {
int n = stones.length;
parent = new int[n];
size = new int[n];
for(int i = 0; i<n; i++){
parent[i] = i;
size[i] = 1;
} int count = n; for(int i = 0; i<n; i++){ int [] stone = stones[i]; if(!row.containsKey(stone[0])){
row.put(stone[0], i);
}else{
int p = row.get(stone[0]);
if(!find(i, p)){
union(i, p);
count--;
} } if(!colum.containsKey(stone[1])){
colum.put(stone[1], i);
}else{
int p = colum.get(stone[1]);
if(!find(i, p)){
union(i, p);
count--;
}
} } return n - count;
} private boolean find(int i, int j){
return root(i) == root(j);
} private int root(int i){
while(i != parent[i]){
parent[i] = parent[parent[i]];
i = parent[i];
} return i;
} private void union(int i, int j){
int p = root(i);
int q = root(j);
if(size[p] > size[q]){
parent[q] = p;
size[p] += size[q];
}else{
parent[p] = q;
size[q] += size[p];
}
}
}
LeetCode 947. Most Stones Removed with Same Row or Column的更多相关文章
- 【LeetCode】947. Most Stones Removed with Same Row or Column 解题报告(Python & C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 并查集 日期 题目地址:https://leetco ...
- 【leetcode】947. Most Stones Removed with Same Row or Column
题目如下: On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may h ...
- Leetcode: Most Stones Removed with Same Row or Column
On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may have at ...
- [Swift]LeetCode947. 移除最多的同行或同列石头 | Most Stones Removed with Same Row or Column
On a 2D plane, we place stones at some integer coordinate points. Each coordinate point may have at ...
- Write an algorithm such that if an element in an MxN matrix is 0, its entire row and column is set to 0.
1: /// <summary> 2: /// Write an algorithm such that if an element in an MxN matrix is 0, it ...
- excel小技巧-用于测试用例的编号栏:“获取当前单元格的上一格的值+1”=INDIRECT(ADDRESS(ROW()-1,COLUMN()))+1
编写用例的时候使用,经常修改用例的时候会需要增加.删除.修改条目,如果用下拉更新数值的方式会很麻烦. 1.使用ctrl下拉,增删移动用例的时候,需要每次都去拉,万一列表比较长,会很麻烦 2.使用ROW ...
- Flutter 布局(七)- Row、Column详解
本文主要介绍Flutter布局中的Row.Column控件,详细介绍了其布局行为以及使用场景,并对源码进行了分析. 1. Row A widget that displays its children ...
- params.row[params.column.key] vue h函数 当前单元格 h函数 div 属性 值或数组 render
params.row[params.column.key] vue h函数 当前单元格 h函数 div 属性 值或数组 render
- Flutter 布局类组件:线性布局(Row和Column)
前言 所谓线性布局,即指沿水平或垂直方向排布子组件.Flutter中通过Row和Column来实现线性布局,并且它们都继承自弹性布局(Flex). 接口描述 Row({ Key key, // 表示子 ...
随机推荐
- sql语句修改数据库字段的长度
修改字段的长度 alter table [OtpimizeRoute_Test].[dbo].[T_QueueFile] alter column ListFileName nvarchar()
- CSS3弹性盒布局方式
一.CSS3弹性盒子 弹性盒子是CSS3的一种新布局模式. CSS3 弹性盒( Flexible Box 或 flexbox),是一种当页面需要适应不同的屏幕大小以及设备类型时确保元素拥有恰当的行为的 ...
- Oracle打印输出在控制台
SET SERVEROUTPUT ON --必须有,不然显示不出declare LN_C number(10,0):=0;begin DECLARE LS_STR1 VARCHAR2(200); - ...
- 「UR#5」怎样跑得更快
「UR#5」怎样跑得更快 膜这个您就会了 下面是复读机mangoyang 我们要求 \[ \sum_{j=1}^n \gcd(i,j)^{c-d} j^d x_j=\frac{b_i}{i^d} \] ...
- kie-api介绍和使用
参考:KIE kie在drools jbpm uberfire里广泛被使用,下面对kie-api中的几个重要组件做下简单介绍 maven依赖 <dependency> <groupI ...
- feign.FeignException: status 400 reading
feign.FeignException: status 400 reading : 请求方调用报错: 服务方被调用报错: 用fegin给redis设置缓存时报错,刚好 卡到8k这个临界点 ,就一直报 ...
- tf.concat的用法
import numpy as npimport tensorflow as tfsess=tf.Session()a=np.zeros((1,2,3,4))b=np.ones((1,2,3,4))c ...
- 2019 多益网络java面试笔试题 (含面试题解析)
本人5年开发经验.18年年底开始跑路找工作,在互联网寒冬下成功拿到阿里巴巴.今日头条.多益网络等公司offer,岗位是Java后端开发,因为发展原因最终选择去了多益网络,入职一年时间了,也成为了面 ...
- P2613 【模板】有理数取余 (数论)
题目 P2613 [模板]有理数取余 解析 简单的数论题 发现并没有对小数取余这一说,所以我们把原式化一下, \[(c=\frac{a}{b})\equiv a\times b^{-1}(mod\ p ...
- pandas-03 DataFrame()中的iloc和loc用法
pandas-03 DataFrame()中的iloc和loc用法 简单的说: iloc,即index locate 用index索引进行定位,所以参数是整型,如:df.iloc[10:20, 3:5 ...