每日一练:岛屿数量
题目要求:
给你一个由 '1'
(陆地)和 '0'
(水)组成的的二维网格,请你计算网格中岛屿的数量。
岛屿总是被水包围,并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
此外,你可以假设该网格的四条边均被水包围。
示例 1:
输入:grid = [ ["1","1","1","1","0"], ["1","1","0","1","0"], ["1","1","0","0","0"], ["0","0","0","0","0"] ] 输出:1
示例 2:
输入:grid = [ ["1","1","0","0","0"], ["1","1","0","0","0"], ["0","0","1","0","0"], ["0","0","0","1","1"] ] 输出:3
提示:
m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j]
的值为'0'
或'1'
解法-1 感染 O(NM)
创建一个数组tab,用来记录grid中的1是否已经有归属,如果是0表示没有归属;
再写一个感染递归函数infect,只要发现没有归属的土地就以它为起点"感染"周围的土地,即让它们合并成一个岛屿;
class Solution {
void infect(vector<vector<char>>& grid, vector<vector<int>>& tab, int i, int j)
{
if (i >= grid[0].size() || j >= grid.size())
return;
if (grid[j][i] == '0')
return;
else if(tab[j][i] == 0)
tab[j][i] = 1;
else
return;
// 上下左右都要合并
infect(grid, tab, i + 1, j);
infect(grid, tab, i, j + 1);
infect(grid, tab, i, j-1);
infect(grid, tab, i-1, j );
}
public:
int numIslands(vector<vector<char>>& grid) {
vector<vector<int>> tab;
tab.resize(grid.size());
for (auto& v : tab)
v.resize(grid[0].size(), 0);
int ret = 0;
for (int i = 0; i < grid[0].size(); i++) {
for (int j = 0; j < grid.size(); j++) {
if (grid[j][i] == '1' && tab[j][i] == 0) { // 无归属土地
ret++;
infect(grid, tab, i, j); // 合并周围土地成一块岛屿
}
}
}
return ret;
}
};
空间优化:
其实可以不用tab存储土地的情况,因为如果土地有归属我们可以用一个其他字符表示在原位置,例如我就用‘2’来表示有归属的土地:
class Solution {
void infect(vector<vector<char>>& grid,int i, int j)
{
if (i >= grid[0].size() || j >= grid.size())
return;
if (grid[j][i] == '1') // 无归属土地
grid[j][i] = '2'; // 标记为有归属
else
return;
infect(grid, i + 1, j);
infect(grid, i, j + 1);
infect(grid, i, j-1);
infect(grid, i-1, j );
}
public:
int numIslands(vector<vector<char>>& grid) {
int ret = 0;
for (int i = 0; i < grid[0].size(); i++) {
for (int j = 0; j < grid.size(); j++) {
if (grid[j][i] == '1') { // 无归属土地
ret++;
infect(grid, i, j);
}
}
}
return ret;
}
};
原文地址:https://blog.csdn.net/2303_78095330/article/details/142637516
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