42. Trapping Rain Water

Hard (Khó) C++ 🔗 Xem trên LeetCode

📋 Đề Bài

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

 

Example 1:

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.

Example 2:

Input: height = [4,2,0,3,2,5]
Output: 9

 

Constraints:

  • n == height.length
  • 1 <= n <= 2 * 104
  • 0 <= height[i] <= 105

💻 Lời Giải

C++ 0042-trapping-rain-water.cpp 
class Solution {
public:
    int trap(vector<int>& height) {
        int n = height.size();
        vector<int> left(n), right(n);
        left[0] = height[0];
        for (int i = 1; i < n; ++i) {
            left[i] = max(left[i - 1], height[i]);
        }
        right[n - 1] = height[n - 1];
        for (int i = n - 2; i >= 0; --i) {
            right[i] = max(right[i + 1], height[i]);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += min(left[i], right[i]) - height[i];
        }
        return ans;
    }
};