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42. Trapping Rain Water

Hard

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

Solution

public class Solution {
    public int Trap(int[] height) {
        int l = 0;
        int r = height.Length - 1;
        int res = 0;
        int lowerWall = 0;
        while (l < r) {
            int lVal = height[l];
            int rVal = height[r];
            // If left is smaller than right ptr, make the lower wall the bigger of lVal and its
            // current size
            if (lVal < rVal) {
                // If lVal has gone up, move the lowerWall upp
                lowerWall = System.Math.Max(lVal, lowerWall);
                // Add the water level at current point
                // Calculate this by taking the current value and subtracting it from the lower wall
                // size
                // We know that this is the lower wall because we've already determined that lVal <
                // rVal
                res += lowerWall - lVal;
                // Move left ptr along
                l++;
            } else {
                // Do the same thing, except now we know that the lowerWall is the right side.
                lowerWall = System.Math.Max(rVal, lowerWall);
                res += lowerWall - rVal;
                r--;
            }
        }
        return res;
    }
}

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