42. Trapping Rain Water

Problem

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

Intuition

// TODO

Approach

// TODO

Implementation

class Solution {
    public int trap(int[] height) {
        int l = 0; int r = height.length-1;

        int water = 0;
        while (l < r) {
            if (height[l] <= height[r]) {
                int k = l; // index dau tien height[k] > height[l]
                do {
                    k++;
                    if (height[k] < height[l] && height[k] < height[r]) {
                        water = water + (height[l] - height[k]);
                    }
                } while (k < r && height[k] <= height[l]);
                l = k;
            } else { // height[l] > height[r]
                int k = r; // index dau tien height[k] > height[r]
                do {
                    k--;
                    if (height[k] < height[l] && height[k] < height[r]) {
                        water = water + (height[r] - height[k]);
                    }
                } while (l < k && height[k] <= height[r]);
                r = k;
            }
        }

        return water;
    }
}