【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/
题目地址:https://leetcode.com/problems/most-stones-removed-with-same-row-or-column/description/
题目描述
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
题目大意
在二维坐标的整数坐标点上,有一些石头,如果一个石头和另外一个石头的横坐标或者纵坐标相等,那么认为他们是有链接的。我们每次取一个和别人有链接的石头,问最终能取得多少个石头。
解题方法
并查集
这个题翻译一下就是,横或者纵坐标相等的坐标点会互相链接构成一个区域,问总的有多少个独立的区域。结果是总的石头数减去独立区域数。
所以,我们根本不用考虑太多,只需要统计有多少区域即可。这个方法最简单的就是并查集。
思路是,两重循环,分别判断石头两两之间是否有链接,如果有链接,那么把他们组成同一个区域。这样的优点是我们只需要和石头等长的数组放每个的parent即可。最后统计最后的区域中-1的数量就是独立区域的个数。石头个数减去独立区域数即可。
python代码如下,可惜超时了。
class Solution(object):
def removeStones(self, stones):
"""
:type stones: List[List[int]]
:rtype: int
"""
N = len(stones)
self.map = [-1] * N
for i in range(N):
for j in range(i + 1, N):
if stones[i][0] == stones[j][0] or stones[i][1] == stones[j][1]:
self.union(i, j)
res = N
print(self.map)
for i in range(N):
if self.map[i] == -1:
res -= 1
return res
def find(self, x):
return x if self.map[x] == -1 else self.find(self.map[x])
def union(self, x, y):
fx = self.find(x)
fy = self.find(y)
if fx != fy:
self.map[fx] = fy
这个版本的C++可以做通过,说明了C++速度的优越性。
class Solution {
public:
int removeStones(vector<vector<int>>& stones) {
if(stones.size() <= 1) return 0;
int N = stones.size();
vector<int> p(N, -1);
for (int i = 0; i < N; ++i){
for(int j = i + 1; j < N; ++j){
if (stones[i][0] == stones[j][0] || stones[i][1] == stones[j][1]){
u(p, i, j);
}
}
}
int res = N;
for(auto e: p) if(e == -1) --res;
return res;
}
private:
int f(vector<int> &p, int x){
if(p[x] == -1) return x;
return f(p, p[x]);
}
void u(vector<int> &p, int x, int y){
int px = f(p, x);
int py = f(p, y);
if(px != py){
p[px] = py;
}
}
};
其实上面这个版本可以做优化,我们不用对石头进行两两判断,而是对他们的横纵坐标同等看待。怎么区分横纵坐标呢?使用的方法是把纵坐标+10000,这样行的索引没变,纵坐标的范围跑到了后面去了。
这个做法的思路是,一个坐标其实就是把横纵坐标对应的两个区域进行了链接。所以,只需要对stones进行一次遍历把对应的区域链接到一起即可。在完成链接之后,我们最后统计一下有多少个独立的区域,需要使用set+find。
Python代码如下:
class Solution(object):
def removeStones(self, stones):
"""
:type stones: List[List[int]]
:rtype: int
"""
N = len(stones)
self.map = [-1] * 20000
for x, y in stones:
self.union(x, y + 10000)
count = set()
return N - len({self.find(x) for x, y in stones})
def find(self, x):
return x if self.map[x] == -1 else self.find(self.map[x])
def union(self, x, y):
fx = self.find(x)
fy = self.find(y)
if fx != fy:
self.map[fx] = fy
C++代码如下:
class Solution {
public:
int removeStones(vector<vector<int>>& stones) {
if(stones.size() <= 1) return 0;
int N = stones.size();
vector<int> p(20000, -1);
for(auto stone : stones){
u(p, stone[0], stone[1] + 10000);
}
set<int> parents;
for(auto stone : stones){
parents.insert(f(p, stone[0]));
}
return N - parents.size();
}
private:
int f(vector<int> &p, int x){
if(p[x] == -1) return x;
return f(p, p[x]);
}
void u(vector<int> &p, int x, int y){
int px = f(p, x);
int py = f(p, y);
if(px != py){
p[px] = py;
}
}
};
日期
2018 年 11 月 24 日 —— 周日开始!一周就过去了~
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