题意

题目

思路

我一开始想的时候只考虑到一个结点周围的边界的情况，并没有考虑到边界的高度其实影响到所有的结点盛水的高度。

我们可以发现，中间是否能够盛水取决于边界是否足够高于里面的高度，所以这必然是一个从外到内，从小到大的一个过程。因为优先队列必然首先访问的是边界中最小的高度，如果周围小于这个高度那么必然是存在可以盛水的地方，就算是没有也没有任何的损失，只是更新了高度，但是队列依然会从高度小的结点开始访问；

实现

typedef struct node {int x, y;node(int x, int y) : x(x), y(y) {}
} Node;class Solution {
public:/**[1,4,3,1,3,2],[3,2,1,3,2,4],[2,3,3,2,3,1]发现：四条边都不可以存储水，，也就是在这个二维数组中才能存储水从第二列开始，去找四周里面大于它自己的第一个数（前提是三面或者两面是大于它的），然后减去自己*/int trapRainWater(vector<vector<int>>& heightMap) {if (heightMap.size() == 0) {return 0;}else if (heightMap.size() < 3 || heightMap[0].size() < 3) {return 0;}int row = heightMap.size();int col = heightMap[0].size();queue<Node*> queues;vector<vector<int>> visited(row, vector<int>(col, 0));Node *start = new Node(1, 1);visited[1][1] = 1;queues.push(start);auto state_center = [&](int x, int y) {if (x >= 1 && x <= heightMap.size()-2 && y >= 1 && y <= heightMap[0].size()-2) {return true;}return false;};//上下左右vector< vector<int>> layouts = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};int sum = 0;while (!queues.empty()) {Node* node = queues.front();queues.pop();// 取中间的值if (state_center(node->x, node->y)) {vector<int> priority;bool isfinished = true;// 找周围四个点for (int i = 0; i < layouts.size(); i++) {vector<int> temp = layouts[i];int newx = node->x + temp[0];int newy = node->y + temp[1];// 不符合条件if (heightMap[newx][newy] > heightMap[node->x][node->y] && (newx == 0 || newx == heightMap.size()-1) &&(newy == 0 || newy == heightMap[0].size()-1)) {isfinished = false;break;}else if (newx == 0 || newx == heightMap.size()-1 || newy == 0 || newy == heightMap[0].size()-1) {priority.push_back(heightMap[newx][newy]);continue;}Node *newnode = new Node(newx, newy);if (!visited[newx][newy]) {queues.push(newnode);visited[newx][newy] = 1;}}if (isfinished) {sort(priority.begin(), priority.end());int val = heightMap[node->x][node->y];for (int i = 0; i < priority.size(); i++) {if (priority[i] >= val) {sum += priority[i] - val;break;}}}}}return sum;}/*** 存放水的高度取决于周围的最小高度，BFS的思想* 首先存取四面的高度，用优先队列去存储，保证取出来的一定是队列中的最小值* 取较大的高度，如果存在没访问过并且小于当前最小高度的，则计算盛水高度，并且将该结点设置成已访问** @param heightMap* @return*/int trapRainWater2(vector<vector<int>>& heightMap) {if (heightMap.size() == 0) {return 0;}int row = heightMap.size();int col = heightMap[0].size();struct cmp {bool operator()(pair<int, int> a, pair<int, int> b) {if (a.first > b.first) {return true;}else {return false;}}};priority_queue<pair<int, int>, vector<pair<int, int>>, cmp> queue;vector<vector<int>> visited(row, vector<int>(col, 0));// 将外围的高度加入到队列中，找出最小值for (int i = 0; i < row; i++) {for (int j = 0; j < col; j++) {if (i == 0 || i == row-1 || j == 0 || j == col-1) {queue.push(make_pair(heightMap[i][j], i * col + j));visited[i][j] = 1;}}}vector< vector<int>> layouts = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};int maxHeight = INT_MIN, sum = 0;while (!queue.empty()) {pair<int, int> content = queue.top();queue.pop();int height = content.first;int x = content.second / col;int y = content.second % col;maxHeight = max(maxHeight, height);for (auto temp : layouts) {int newx = x + temp[0];int newy = y + temp[1];if (newx < 0 || newx >= row || newy < 0 || newy >= col || visited[newx][newy]) {continue;}if (heightMap[newx][newy] < maxHeight) {sum += maxHeight - heightMap[newx][newy];}visited[newx][newy] = 1;queue.push(make_pair(heightMap[newx][newy], newx * col + newy));}}return sum;}void test() {vector< vector<int>> water = {{12,13,1,12},{13,4,13,12},{13,8,10,12},{12,13,12,12},{13,13,13,13}};int result = this->trapRainWater2(water);cout << "sum:" << result << endl;}
};

参考文章