LeetCode 42. Trapping Rain Water

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

For example,
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

To solve the problem. the key point to find the maxHeight  index.

#include <vector>#include <iostream>#include <climits>using namespace std;/*  Given n non-negative integers representing an elevation map where the width of each bar is 1.  compute how much water it is able to trap after raining.  For example:  Given [0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1], return 6.*/int trap(vector<int>& height) {  int maxHeight = 0;  int maxIndex = 0;  for(int i = 0; i < height.size(); ++i) {    if(maxHeight < height[i]) {      maxHeight = height[i];      maxIndex = i;    }  }  int water = 0;  int maxSoFar = 0;  for(int i = 0; i < <strong>maxIndex</strong>; ++i) {    if(height[i] < maxSoFar) water += (maxSoFar - height[i]); // less then maxSoFar can accumulate water.    else maxSoFar = height[i];  }  maxSoFar = 0;  for(int i = height.size() - 1; i > <strong>maxIndex</strong>; --i) {    if(height[i] < maxSoFar) water += (maxSoFar - height[i]);  // less then maxSoFar can accumulate water.    else maxSoFar = height[i];  }  return water;}int main(void) {  vector<int> nums{0, 1, 0, 3, 1, 0, 1, 3, 2, 1, 2, 1};  cout << trap(nums) << endl;}

Another interesting variation! Trapping rain water in matrix

/*  Suppose given a matrix  {{0, 2, 5},   {5, 4, 5},   {5, 5, 5}};  In this matrix, the max water can be trapped is: 0;*/#include "header.h"using namespace std;class Solution {private:  vector< vector<int> > matrix;public:  Solution(vector< vector<int> >& input) {    matrix = input;  }  int trappWaterinMatrix() {    if(matrix.size() == 0 || matrix[0].size() == 0) return 0;    int rows = matrix.size();    int cols = matrix[0].size();    struct posIndex {      int value;      int row;      int col;      posIndex(int v, int r, int c) : value(v), row(r), col(c) {}      bool operator < (posIndex b) const {        return value > b.value;      }    };   priority_queue<posIndex, vector<posIndex> > minHeap;    vector< vector<bool> > visited(rows, vector<bool>(cols, false));    // we need to first get the minimum value from the border.    // first row and last row    for(int i = 0; i < cols; ++i) {      minHeap.push(posIndex(matrix[0][i], 0, i));      visited[0][i] = true;      minHeap.push(posIndex(matrix[rows - 1][i], rows - 1, i));      visited[rows - 1][i] = true;    }    // first column and last column    for(int i = 1; i < rows; ++i) {      minHeap.push(posIndex(matrix[i][0], i, 0));      visited[i][0] = true;      minHeap.push(posIndex(matrix[i][cols -1], i, cols -1));      visited[i][cols - 1] = true;    }    int waterTrapped = 0;    // bfs traversal the matrix to caculate the trapped water.    vector< vector<int> > dir{{0, 1}, {0, -1}, {-1, 0}, {1, 0}};    while(!minHeap.empty()) {      auto curr = minHeap.top(); minHeap.pop();      for(int i = 0; i < dir.size(); ++i) {        posIndex next(0, 0, 0);        next.row = curr.row + dir[i][0];        next.col = curr.col + dir[i][1];        if(next.row >= 0 && next.row < rows && next.col >= 0 && next.col < cols && !visited[next.row][next.col]) {          visited[next.row][next.col] = true;          next.value = max(curr.value, matrix[next.row][next.col]);          minHeap.push(next);          waterTrapped += max(0, curr.value - matrix[next.row][next.col]);        }      }    }    return waterTrapped;  }};