LeetCode【42. Catching rainwater】

I don't like to wait for others, and I don't like to be waited for by others.

Given  n a non-negative integer representing  1 the height map of each column with a width of , calculate how much rainwater the columns arranged in this way can catch after it rains.

Example 1:

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
 Output: 6
 Explanation: The above is composed of the array [0,1,0,2,1,0 ,1,3,2,1,2,1] represents the height map. In this case, 6 units of rainwater can be received (the blue part represents rainwater). 

Example 2:

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

hint:

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

Number of passes 773.8K Number of submissions 1.2M Pass rate 63.1%

The solution to the rainwater problem also adopts the double-pointer method

public int trap(int[] height) {
    int left = 0;
    int right = height.length - 1;
    int leftMax = 0;
    int rightMax = 0;
    int result = 0;

    while (left < right) {
        if (height[left] < height[right]) {
            if (height[left] >= leftMax) {
                leftMax = height[left];
            } else {
                result += leftMax - height[left];
            }
            left++;
        } else {
            if (height[right] >= rightMax) {
                rightMax = height[right];
            } else {
                result += rightMax - height[right];
            }
            right--;
        }
    }

    return result;
}