200 / 130 Number of Islands / Surrounded Regions
Total Accepted: 48411
Total Submissions: 171609
Difficulty: Medium
Given a 2d grid map of '1's (land) and '0's (water),
count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically.
You may assume all four edges of the grid are all surrounded by water.
Example 1:
11110
11010
11000
00000
Answer: 1
Example 2:
11000
11000
00100
00011
Answer: 3
分析：
这是网上流传最广的深度优先解法，确实漂亮，通俗易理解。
class Solution {
public:
int numIslands(vector>& grid) {
int result=0;
if(grid.empty() || grid[0].empty())
return result;
int rows=grid.size();
int cols=grid[0].size();
for(int i=0;i>& grid,int i,int j,int rows,int cols)
{
grid[i][j]='0';
if(i > 0 && grid[i-1][j] == '1')//上边置0
dfs(grid,i-1,j,rows,cols);
if(i < rows-1 && grid[i+1][j] == '1')//下边同理
dfs(grid,i+1,j,rows,cols);
if(j > 0 && grid[i][j-1] == '1')//左边
dfs(grid,i,j-1,rows,cols);
if(j < cols-1 && grid[i][j+1] == '1')//右边
dfs(grid,i,j+1,rows,cols);
}
}; 学习别人的并查集解决问题，漂亮：
class Solution {
public:
int find(vector & parents, int index) {//寻找index位置的最高祖先（也是一个位置）
if(parents[index]==-1)
return index;
return find(parents, parents[index]);
}
void merge(vector & parents, int a, int b){//合并
int parent_a = find(parents, a);
int parent_b = find(parents, b);
if(parent_a!=parent_b){
parents[parent_b]=parent_a;//set the new 's parent to be old one.
total--;
}
}
int numIslands(vector>& grid) {
int rows = grid.size();
if (rows==0) return 0;
int cols = grid[0].size();
vectorparents(rows*cols,-1);
total=rows*cols; //随后减少
int waters=0;
for(int i=0;i0 && grid[i-1][j]=='1')
merge(parents, (i-1)*cols+j, index);
if(j>0 && grid[i][j-1]=='1')
merge(parents, i*cols+j-1, index);
}
}
}
return total-waters;
}
private:
int total;
};
Total Accepted: 54885
Total Submissions: 336458
Difficulty: Medium
Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.
A region is captured by flipping all 'O's into 'X's in that surrounded region.
For example,
X X X X X O O X X X O X X O X X
After running your function, the board should be:
X X X X X X X X X X X X X O X X
分析：
拖上一题的福，200. Number of Islands，这一题思路比较明显。
以下答案Runtime Error，查了一下网络，应该是栈溢出了！
class Solution {
public:
void solve(vector>& board) {
int rows=board.size();
if(rows==0 || rows==1)
return;
int cols=board[0].size();
for(int i=0;i>& board,int i,int j,int rows,int cols)
{
board[i][j]='D';//将其改成D，区别于被包围的O，便于在最后两层循环中区别那些该被改成X
if(i < rows-1 && board[i+1][j] == 'O')//下边
dfs(board,i+1,j,rows,cols);
if(j < cols-1 && board[i][j+1] == 'O')//右边
dfs(board,i,j+1,rows,cols);
if(i > 0 && board[i-1][j] == 'O')//上边
dfs(board,i-1,j,rows,cols);
if(j > 0 && board[i][j-1] == 'O')//zuo边
dfs(board,i,j-1,rows,cols);
}
}; 学习别人的算法：
广度优先算法
/**------------------------------------
* 日期：2015-02-06
* 作者：SJF0115
* 题目: 130.Surrounded Regions
* 网址：https://oj.leetcode.com/problems/surrounded-regions/
* 结果：AC
* 来源：LeetCode
* 博客：
---------------------------------------**/
#include
#include
#include
#include
#include
using namespace std;
class Solution {
public:
void solve(vector > &board) {
int row = board.size();
if(row == 0){
return;
}//if
int col = board[0].size();
// 都够不成围绕
if(row <= 2 || col <= 2){
return;
}//if
// 行
for(int i = 0;i < col;++i){
// 第一行
BFS(board,row,col,0,i);
// 最后一行
BFS(board,row,col,row-1,i);
}//for
// 列
for(int j = 0;j < row;++j){
// 第一列
BFS(board,row,col,j,0);
// 最后一列
BFS(board,row,col,j,col-1);
}//for
for(int i = 0;i < row;++i){
for(int j = 0;j < col;j++){
// 不可以从外界走通的o
if(board[i][j] == 'O'){
board[i][j] = 'X';
}//if
// 可以从外界走通的o
else if(board[i][j] == '.'){
board[i][j] = 'O';
}
}//for
}//for
}
private:
// row 行数 col 列数 x ,y 当前board位置
void BFS(vector> &board,int row,int col,int x,int y){
queue > q;
Pass(board,row,col,x,y,q);
while(!q.empty()){
pair point = q.front();
q.pop();
x = point.first;
y = point.second;
// left
Pass(board,row,col,x,y+1,q);
// right
Pass(board,row,col,x,y-1,q);
// up
Pass(board,row,col,x-1,y,q);
// down
Pass(board,row,col,x+1,y,q);
}//while
}
// 四边判断是否走通
void Pass(vector> &board,int row,int col,int x,int y,queue > &q){
// 边界条件以及遇到o才能走通
if(x < 0 || x >= row || y < 0 || y >= col || board[x][y] != 'O'){
return;
}//if
// 标记可从外界走通的o
board[x][y] = '.';
// 入队列
q.push(make_pair(x,y));
}
};
int main(){
Solution s;
/*vector > board = {
{'X','X','X','X'},
{'X','O','O','X'},
{'X','X','O','X'},
{'X','O','X','X'}
};*/
vector > board = {
{'X','X','X'},
{'X','O','X'},
{'X','X','X'}
};
s.solve(board);
// 输出
for(int i = 0;i < board.size();i++){
for(int j = 0;j < board[i].size();j++){
cout< 注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！
原文地址：http://blog.csdn.net/ebowtang/article/details/51636977
原作者博客：http://blog.csdn.net/ebowtang
本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895