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200.岛屿数量

1. BFS

遍历与BFS。时间复杂度O(mn),空间复杂度O(mn)(visited数组)。

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class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
int m = grid.size();
if (m == 0)
return 0;

int n = grid[0].size();
if (n == 0)
return 0;

vector<vector<bool> > visited(m, vector<bool>(n, false));
int mvi[4] = {-1, 1, 0, 0};
int mvj[4] = {0, 0, -1, 1};
queue<pair<int, int> > q;
int ans = 0;

for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (!visited[i][j] && grid[i][j] == '1') {
q.push(make_pair(i, j));
visited[i][j] = true;
while (!q.empty()) {
pair<int, int> loc = q.front();
q.pop();
int loci = loc.first, locj = loc.second;
for (int k = 0; k < 4; k++) {
int curi = loci + mvi[k], curj = locj + mvj[k];
if (curi >= 0 && curi < m && curj >= 0 && curj < n && grid[curi][curj] == '1' && !visited[curi][curj]) {
q.push(make_pair(curi, curj));
visited[curi][curj] = true;
}
}
}
ans++;
}
}
}

return ans;
}
};